Conjectured value of a harmonic sum $sum_n=1^inftyleft(H_n-,2H_2n+H_4nright)^2$Infinite Series $sum_n=1^inftyfracH_nn^32^n$Sum of Squares of Harmonic NumbersA closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Closed form for $sum_n=1^inftyfrac(-1)^n n^4 H_n2^n$Closed form for $sum_n=1^inftyfrac(-1)^n n^a H_n2^n$A closed form of the series $sum_n=1^infty fracH_n^2-(gamma + ln n)^2n$Infinite Series $sum_n=1^inftyfracH_nn^32^n$A conjectured result for $sum_n=1^inftyfrac(-1)^n,H_n/5n$Strategies for evaluating sums $sum_n=1^infty fracH_n^(m)z^nn$How to evaluate $lim_nrightarrow infty nleft [ widetildeH_n-H_2n+H_n right ]$?What's about $sum_n=1^infty frace^H_nlog H_nn^3$, where $H_n$ is the nth harmonic number?A closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Justify an approximation of $-sum_n=2^infty H_nleft(frac1zeta(n)-1right)$, where $H_n$ denotes the $n$th harmonic number
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Conjectured value of a harmonic sum $sum_n=1^inftyleft(H_n-,2H_2n+H_4nright)^2$
Infinite Series $sum_n=1^inftyfracH_nn^32^n$Sum of Squares of Harmonic NumbersA closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Closed form for $sum_n=1^inftyfrac(-1)^n n^4 H_n2^n$Closed form for $sum_n=1^inftyfrac(-1)^n n^a H_n2^n$A closed form of the series $sum_n=1^infty fracH_n^2-(gamma + ln n)^2n$Infinite Series $sum_n=1^inftyfracH_nn^32^n$A conjectured result for $sum_n=1^inftyfrac(-1)^n,H_n/5n$Strategies for evaluating sums $sum_n=1^infty fracH_n^(m)z^nn$How to evaluate $lim_nrightarrow infty nleft [ widetildeH_n-H_2n+H_n right ]$?What's about $sum_n=1^infty frace^H_nlog H_nn^3$, where $H_n$ is the nth harmonic number?A closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Justify an approximation of $-sum_n=2^infty H_nleft(frac1zeta(n)-1right)$, where $H_n$ denotes the $n$th harmonic number
$begingroup$
There is a known asymptotic expansion of harmonic numbers $H_n$ for $ntoinfty$:
$$beginalignH_n&=gamma+ln n+sum_k=1^inftyleft(-fracB_kkcdot n^kright)\
&=gamma+ln n+frac12n-frac112n^2+frac1120n^4-frac1252n^6,+,dots,endaligntag1$$
where $B_k$ are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for $O(n^-1)$ term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
$$sum_k=1^inftyleft(H_n-,2H_2n+H_4n-frac18nright)=frac18-fracpi16.tag2$$
Rather than compensating for $O(n^-1)$ term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
$$sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)=frac3pi16-fracpi4sqrt2-fracln28.tag3$$
Generalizing, we can consider two families of series:
$$mathcal A_m=sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)^m,tag4$$
$$mathcal B_m=sum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^m,tag5$$
and try to evaluate them in a closed form.
So far I have the following conjectured result:
$$largesum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^2stackrelnormalsizecolorgray?=fracpi8-fracpi16,ln2-fracpi^296+frac316,ln^22-fracG4,tag$diamond$$$
where $G$ is the Catalan constant.
Could you please help me to prove this result and, possibly, find other values of $mathcal A_m,mathcal B_m$?
calculus sequences-and-series closed-form conjectures harmonic-numbers
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
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This question has an open bounty worth +50
reputation from TheSimpliFire ending ending at 2019-03-17 09:11:13Z">in 4 days.
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add a comment |
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There is a known asymptotic expansion of harmonic numbers $H_n$ for $ntoinfty$:
$$beginalignH_n&=gamma+ln n+sum_k=1^inftyleft(-fracB_kkcdot n^kright)\
&=gamma+ln n+frac12n-frac112n^2+frac1120n^4-frac1252n^6,+,dots,endaligntag1$$
where $B_k$ are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for $O(n^-1)$ term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
$$sum_k=1^inftyleft(H_n-,2H_2n+H_4n-frac18nright)=frac18-fracpi16.tag2$$
Rather than compensating for $O(n^-1)$ term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
$$sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)=frac3pi16-fracpi4sqrt2-fracln28.tag3$$
Generalizing, we can consider two families of series:
$$mathcal A_m=sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)^m,tag4$$
$$mathcal B_m=sum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^m,tag5$$
and try to evaluate them in a closed form.
So far I have the following conjectured result:
$$largesum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^2stackrelnormalsizecolorgray?=fracpi8-fracpi16,ln2-fracpi^296+frac316,ln^22-fracG4,tag$diamond$$$
where $G$ is the Catalan constant.
Could you please help me to prove this result and, possibly, find other values of $mathcal A_m,mathcal B_m$?
calculus sequences-and-series closed-form conjectures harmonic-numbers
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
$endgroup$
This question has an open bounty worth +50
reputation from TheSimpliFire ending ending at 2019-03-17 09:11:13Z">in 4 days.
One or more of the answers is exemplary and worthy of an additional bounty.
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Some possibly related questions are linked from here.
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– Piotr Shatalin
Aug 22 '15 at 18:18
3
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BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
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– Vladimir Reshetnikov
Aug 22 '15 at 19:07
2
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And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
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– Vladimir Reshetnikov
Aug 22 '15 at 19:43
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One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
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– Redundant Aunt
Aug 22 '15 at 19:56
4
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One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
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– Vladimir Reshetnikov
Aug 22 '15 at 22:14
add a comment |
$begingroup$
There is a known asymptotic expansion of harmonic numbers $H_n$ for $ntoinfty$:
$$beginalignH_n&=gamma+ln n+sum_k=1^inftyleft(-fracB_kkcdot n^kright)\
&=gamma+ln n+frac12n-frac112n^2+frac1120n^4-frac1252n^6,+,dots,endaligntag1$$
where $B_k$ are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for $O(n^-1)$ term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
$$sum_k=1^inftyleft(H_n-,2H_2n+H_4n-frac18nright)=frac18-fracpi16.tag2$$
Rather than compensating for $O(n^-1)$ term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
$$sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)=frac3pi16-fracpi4sqrt2-fracln28.tag3$$
Generalizing, we can consider two families of series:
$$mathcal A_m=sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)^m,tag4$$
$$mathcal B_m=sum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^m,tag5$$
and try to evaluate them in a closed form.
So far I have the following conjectured result:
$$largesum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^2stackrelnormalsizecolorgray?=fracpi8-fracpi16,ln2-fracpi^296+frac316,ln^22-fracG4,tag$diamond$$$
where $G$ is the Catalan constant.
Could you please help me to prove this result and, possibly, find other values of $mathcal A_m,mathcal B_m$?
calculus sequences-and-series closed-form conjectures harmonic-numbers
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
$endgroup$
There is a known asymptotic expansion of harmonic numbers $H_n$ for $ntoinfty$:
$$beginalignH_n&=gamma+ln n+sum_k=1^inftyleft(-fracB_kkcdot n^kright)\
&=gamma+ln n+frac12n-frac112n^2+frac1120n^4-frac1252n^6,+,dots,endaligntag1$$
where $B_k$ are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for $O(n^-1)$ term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
$$sum_k=1^inftyleft(H_n-,2H_2n+H_4n-frac18nright)=frac18-fracpi16.tag2$$
Rather than compensating for $O(n^-1)$ term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
$$sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)=frac3pi16-fracpi4sqrt2-fracln28.tag3$$
Generalizing, we can consider two families of series:
$$mathcal A_m=sum_n=1^infty,(-1)^n,Big(H_n-,2H_2n+H_4nBig)^m,tag4$$
$$mathcal B_m=sum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^m,tag5$$
and try to evaluate them in a closed form.
So far I have the following conjectured result:
$$largesum_n=1^inftyBig(H_n-,2H_2n+H_4nBig)^2stackrelnormalsizecolorgray?=fracpi8-fracpi16,ln2-fracpi^296+frac316,ln^22-fracG4,tag$diamond$$$
where $G$ is the Catalan constant.
Could you please help me to prove this result and, possibly, find other values of $mathcal A_m,mathcal B_m$?
calculus sequences-and-series closed-form conjectures harmonic-numbers
calculus sequences-and-series closed-form conjectures harmonic-numbers
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
asked Aug 22 '15 at 17:46
Vladimir ReshetnikovVladimir Reshetnikov
24.5k5120235
24.5k5120235
This question has an open bounty worth +50
reputation from TheSimpliFire ending ending at 2019-03-17 09:11:13Z">in 4 days.
One or more of the answers is exemplary and worthy of an additional bounty.
This question has an open bounty worth +50
reputation from TheSimpliFire ending ending at 2019-03-17 09:11:13Z">in 4 days.
One or more of the answers is exemplary and worthy of an additional bounty.
$begingroup$
Some possibly related questions are linked from here.
$endgroup$
– Piotr Shatalin
Aug 22 '15 at 18:18
3
$begingroup$
BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:07
2
$begingroup$
And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:43
$begingroup$
One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
$endgroup$
– Redundant Aunt
Aug 22 '15 at 19:56
4
$begingroup$
One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
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– Vladimir Reshetnikov
Aug 22 '15 at 22:14
add a comment |
$begingroup$
Some possibly related questions are linked from here.
$endgroup$
– Piotr Shatalin
Aug 22 '15 at 18:18
3
$begingroup$
BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:07
2
$begingroup$
And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:43
$begingroup$
One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
$endgroup$
– Redundant Aunt
Aug 22 '15 at 19:56
4
$begingroup$
One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 22:14
$begingroup$
Some possibly related questions are linked from here.
$endgroup$
– Piotr Shatalin
Aug 22 '15 at 18:18
$begingroup$
Some possibly related questions are linked from here.
$endgroup$
– Piotr Shatalin
Aug 22 '15 at 18:18
3
3
$begingroup$
BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:07
$begingroup$
BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:07
2
2
$begingroup$
And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:43
$begingroup$
And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:43
$begingroup$
One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
$endgroup$
– Redundant Aunt
Aug 22 '15 at 19:56
$begingroup$
One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
$endgroup$
– Redundant Aunt
Aug 22 '15 at 19:56
4
4
$begingroup$
One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 22:14
$begingroup$
One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
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– Vladimir Reshetnikov
Aug 22 '15 at 22:14
add a comment |
2 Answers
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So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $displaystyle H_n-2H_2n+H_4n=int_0^1 fracx^2n-x^4n1+xdx$.
I noticed that $H_n-2H_2n+H_4n=H_4n-H_2n-(H_2n-H_n)=H_4n^--H_2n^-$, where $H_n^-=sum_k=1^n frac(-1)^k+1k$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=frac1-x1+x$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -lnsin x=ln2+sum_n=1^infty fraccos(2nx)n ,-lncos x=ln2+sum_n=1^infty frac(-1)^ncos(2nx)n tag1$
$$ int_0^fracpi2 cos x cos(nx)dx=begincases fracpi4 &n=1\0 &n ,,textodd\ frac(-1)^1+n/2n^2-1 &n ,,texteven endcases tag2$$
$$ int_0^1 fracln(1-x)a+xdx=-operatornameLi_2left(frac1a+1right)tag3$$
Starting,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=sum_n=0^inftyint_0^1int_0^1frac(x^2n-x^4n)(u^2n-u^4n)(1+x)(1+u)dxdu
\=smallint_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)-2int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^4)+int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^4)
\=I_22-2I_24+I_44$$
Computing $I_22$.
Substitute $u=fracyx$ ,change the order of integration, evaluate the inner integral, and substitue $t=frac1-x1+x$ to get
$$beginalign I_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)=int_0^1int_0^xfracdydx(1+x)(x+y)(1-y^2)
\=int_0^1 frac11-y^2int_1^y fracdx(1+x)(x+y) dy=int_0^1 fraclnleft(frac(1+x)^24xright)(1+x)(1-x^2),dx
\=frac-14int_0^1 frac(1+t)t^2ln(1-t^2)dt=-frac14int_0^1fracln(1-t^2)t^2dt-frac14int_0^1fracln(1-t^2)tdt
\=frac14sum_n=0^infty frac1(n+1)(2n+1)+frac14sum_n=0^infty frac1(n+1)(2n+2)=fracln22+fracpi^248.endalign$$
Computing $I_44$.
Start the same as with $I_22$ to get $displaystyle I_44=int_0^1 fraclnleft(frac(1+x)^24xright)(1-x)(1-x^4),dx=frac-18int_0^1 fracln(1-t^2)t^2(1+t^2)(1+t)^3dt$.
We can calculate these integrals:
$$beginalign int_0^1 fracln(1-x^2)1+x^2dx=int_0^1 fracln(1+x)1+x^2dx+int_0^1 fracln(1-x)1+x^2dx tag4
\=int_0^1 fracln(1+x)1+x^2dx +int_0^1 fraclnleft(frac2t1+tright)1+t^2dt
\=fracpi4ln2+sum_n=0^infty (-1)^nint_0^1ln(t) t^2ndt=fracpi4ln2-G. endalign$$
$$beginalign int_0^1 fracln(1-x^2)x^2(1+x^2)dx=int_0^1 fracln(1-x^2)x^2dx-int_0^1 fracln(1-x^2)1+x^2dx tag5
\=-sum_n=0^infty frac1n+1int_0^1 x^2ndx-fracpi4ln2+G=G-fracpi4ln2-2ln2.endalign$$
$$beginalign int_0^1 fracxln(1-x^2)1+x^2dx=frac12int_0^1 fracln(1-x)1+xdx tag6
\=-frac12 operatornameLi_2left(frac12right)=fracln^2 24-fracpi^224.endalign$$
$$beginalign int_0^1 fracln(1-x^2)x(1+x^2)dx=int_0^1 fracln(1-x^2)xdx-int_0^1 fracxln(1-x^2)1+x^2dx tag7
\=-sum_n=0^inftyfrac1n+1int_0^1 x^2n+1dx-fracln^2 24+fracpi^224=-fracpi^224-fracln^2 24.endalign$$
Altogether,
$$I_44=frac-18int_0^1 fracln(1-x^2)x^2(1+x^2)(1+3x+3x^2+x^3)dx
\=-fracpi16ln2+fracln24+fracln^2 216+fracpi^248+fracG4.$$
Computing $I_24$.
Substitute $u=fracyx^2$, change the order of integration, let $yto y^2$, evaluate the inner integral,and substitue $t=frac1-x1+x$:
$$beginalign* I_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^2)=int_0^1int_0^x^2 fracdydx(1+x)(x^2+y)(1-y^2)
\=int_0^1 frac11-y^2int_sqrty^1fracdx(1+x)(x^2+y)dy=2int_0^1fracy1-y^4int_y^1fracdx(1+x)(x^2+y^2)dy
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1-x^4)dx
=I_241-I_242. endalign*$$
Evaulation of $I_241$.
Substitute $t=frac1-x1+x$ to get $displaystyle I_241=frac14int_0^1 fractan^-1(t)t(1+t^2)^2(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=tantheta$:
$$beginalign int_0^1 fractan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetacos^2(theta)dtheta=fracpi^264+fracpi16-frac18.tag8endalign$$
$$beginalign int_0^1 fracxtan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetatanthetacos^2thetadtheta=frac12int_0^fracpi4thetasin(2theta)dtheta=frac18.tag9endalign$$
$$beginalign int_0^1 fracx^2tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^2(theta)dtheta=fracpi^264-fracpi16+frac18.tag10endalign$$
$$beginalign int_0^1 fracx^3tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^3(theta)sectheta,dtheta tag11
\=int_0^fracpi4thetatantheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=fracpi8ln2-frac18+int_0^fracpi4lncostheta ,dtheta
\=fracpi8ln2-frac18-int_0^fracpi4ln2 ,dtheta-sum_n=1^infty frac(-1)^nnint_0^fracpi4cos(2ntheta),dtheta
\=-fracpi8ln2-frac18+frac12sum_n=1^inftyfrac(-1)^n+1n^2sin(fracpi n2)=fracG2-fracpi8ln2-frac18.endalign$$
$$beginalign int_0^1 fractan^-1(x)x(1+x^2)^2dx=int_0^fracpi4thetacos^3(theta)csctheta,dtheta tag12
\=int_0^fracpi4thetacottheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=-frac18-fracpi8ln2-int_0^fracpi4lnsintheta ,dtheta
\=-frac18-fracpi8ln2+int_0^fracpi4ln2 ,dtheta+sum_n=1^inftyfrac1nint_0^fracpi4cos(2ntheta),dtheta
\=-frac18+fracpi8ln2+frac12sum_n=1^infty fracsin(fracpi n2)n^2=fracG2+fracpi8ln2-frac18.endalign$$
Altogether,
$$I_241=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx= frac14int_0^1 fractan^-1(x)x(1+x^2)^2(1+4x+6x^2+4x^3+x^4)dx
\=fracpi^232+frac18+fracG4$$
Evaulation of $I_242$.
Substitute $t=frac1-x1+x$ to get
$$ I_242=frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)^2(1-t^2)(1+t)^2dt
\=frac12int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)(1+t^2)^2(1-t^2)dt+frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)(1-t^2)dt
\=frac12int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx-frac14int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx\+frac18int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx-frac14int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx
$$
Calculating these integrals:
$$beginalign int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx=-frac12int_0^1 fraclnleft(frac1-x1+xright)sqrtx(1+x)frac1-x1+xdx tag13
\=-frac12int_0^1fractln tsqrt1-t^2dt=-frac18int_0^1fracln tsqrt1-tdt=-frac18int_0^1 t^-1/2ln(1-t)dt
\=frac14sum_n=0^infty frac1(n+1)(2n+3)=frac12-fracln22.endalign$$
$$beginalign int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1+x)(1-x)x(1+x)dx tag14
\=frac12int_0^1fracln(1+x)xdx-int_0^1fracln(1+x)1+xdx
\=frac12sum_n=0^inftyfrac(-1)^n+1n+1int_0^1 x^n dx-frac12ln^2(1+x)bigg_0^1=fracpi^224-fracln^2 22.endalign$$
$$beginalign int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1-x)xdx-int_0^1fracln(1-x)1+xdx tag15
\=-fracpi^212-left(fracln^2 22-fracpi^212right)=-fracln^2 22.endalign$$
$$
int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx=-2int_0^fracpi4cos^2(theta)(1-tan^2theta)lncostheta,dtheta tag16
\=-2int_0^fracpi4cos(2theta)lncostheta,dtheta=2ln2int_0^fracpi4cos(2theta)dtheta+sum_n=1^infty frac(-1)^nnint_0^fracpi2costheta cos(ntheta)dtheta
\=ln2-fracpi4+sum_n=1^infty frac(-1)^2n2nfrac(-1)^n+1(2n)^2-1=fracln22-fracpi4+frac12.$$
Altogether, $displaystyle I_242=fracpi^2192+fracpi16+fracln^2 216-frac3ln28+frac18$,
leading to $displaystyle I_24=I_241-I_242=frac5pi^2192-fracpi16-fracln^2 216+frac3ln28+fracG4$,
and finally, confirming the conjecture,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=I_22-2I_24+I_44=fracpi8-fracpi16ln2-fracpi^296+frac3ln^2 216-fracG4.$$
I don't know about higher powers. I guess the case $mathcal A_2$ can also be done. If we start the same as with $mathcal B_2$, writing $mathcal A_2=J_22-2J_24+J_44$
we can find that $displaystyle J_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^2u^2)=2I_44-I_22=-fracpi8ln2+fracG2+fracpi^248+fracln^2 28$
$J_44$ can be reduced to $displaystyle =-frac12int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2sqrt2)(x+3-2sqrt2).$
I can get $displaystyle int_0^1 fracln(1-x^2)x(x^4+6x^2+1)dx=-fracpi^212+frac4-3sqrt216operatornameLi_2left(frac2-sqrt24right)+frac4+3sqrt216operatornameLi_2left(frac2+sqrt24right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx=frac1+2sqrt24piln2-fracpi^224-frac14lnleft(frac2+sqrt24right)lnleft(frac2-sqrt24right)
-fracsqrt2+12ImoperatornameLi_2left(frac2+sqrt22+fracisqrt22right)-fracsqrt2-12ImoperatornameLi_2left(frac2-sqrt22+fracisqrt22right)$$
I obtained it by calculating $displaystyle int_0^1 fracln(1+x)x+adx=ln2lnleft(fraca+1a-1right)+operatornameLi_2left(frac21-aright)-operatornameLi_2left(frac11-aright)$,
which together with $(3)$ can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x+adx$, which in turn, through partial fractions, can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x^2+a^2dx$.
Fortunately, things didn't get too ugly as both $3+2sqrt2$ and $3-2sqrt2$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_24$. Starting similarly as with $I_24$,
we have:
$$J_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^4u^2)
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1+x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1+x^4)dx
\=J_241-J_242$$
Through $t=frac1-x1+x$, $J_241$ turns to $displaystyle int_0^1 fractan^-1(x)(1+x^2)(x^4+6x^2+1)(1+x)^4,dx$. I don't have any idea about that yet. Edit 1.
answered Sep 14 '15 at 15:51
nospoonnospoon
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Just my thoughts for now: I would try to exploit Parseval's identity. For first, we have:
$$ H_n-2H_2n+H_4n =int_0^1frac-x^n+2x^2n-x^4n1-x,dx tag1 $$
and:
$$ sum_ngeq 1frac-x^n+2x^2n-x^4n1-xe^niy = frac11-xleft(frac-11-e^iyx+frac21-e^iyx^2+frac-11-e^iyx^4right).tag2$$
The poles of the RHS (as a function of $x$) are located at $xinlefte^-iy,pm e^-iy/2,pm e^-iy/4,pm i e^-iy/4right$.
By using the residue theorem we may compute an explicit representation for:
$$ g(y) = sum_ngeq 1left(H_n-2H_2n+H_4nright)e^niy,tag3 $$
then Parseval's theorem gives:
$$ sum_ngeq 1left(H_n-2H_2n+H_4nright)^2 = frac12piint_-pi^pig(y)g(-y),dy tag4$$
and the resulting integral should be not to difficult to evaluate in terms of dilogarithms.
Another chance may be to apply summation by parts (like I did in this question), but it looks lengthy.
edited Apr 13 '17 at 12:21
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1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
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It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
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– Winther
Sep 1 '15 at 15:26
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@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
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– Jack D'Aurizio
Sep 1 '15 at 15:35
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So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $displaystyle H_n-2H_2n+H_4n=int_0^1 fracx^2n-x^4n1+xdx$.
I noticed that $H_n-2H_2n+H_4n=H_4n-H_2n-(H_2n-H_n)=H_4n^--H_2n^-$, where $H_n^-=sum_k=1^n frac(-1)^k+1k$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=frac1-x1+x$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -lnsin x=ln2+sum_n=1^infty fraccos(2nx)n ,-lncos x=ln2+sum_n=1^infty frac(-1)^ncos(2nx)n tag1$
$$ int_0^fracpi2 cos x cos(nx)dx=begincases fracpi4 &n=1\0 &n ,,textodd\ frac(-1)^1+n/2n^2-1 &n ,,texteven endcases tag2$$
$$ int_0^1 fracln(1-x)a+xdx=-operatornameLi_2left(frac1a+1right)tag3$$
Starting,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=sum_n=0^inftyint_0^1int_0^1frac(x^2n-x^4n)(u^2n-u^4n)(1+x)(1+u)dxdu
\=smallint_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)-2int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^4)+int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^4)
\=I_22-2I_24+I_44$$
Computing $I_22$.
Substitute $u=fracyx$ ,change the order of integration, evaluate the inner integral, and substitue $t=frac1-x1+x$ to get
$$beginalign I_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)=int_0^1int_0^xfracdydx(1+x)(x+y)(1-y^2)
\=int_0^1 frac11-y^2int_1^y fracdx(1+x)(x+y) dy=int_0^1 fraclnleft(frac(1+x)^24xright)(1+x)(1-x^2),dx
\=frac-14int_0^1 frac(1+t)t^2ln(1-t^2)dt=-frac14int_0^1fracln(1-t^2)t^2dt-frac14int_0^1fracln(1-t^2)tdt
\=frac14sum_n=0^infty frac1(n+1)(2n+1)+frac14sum_n=0^infty frac1(n+1)(2n+2)=fracln22+fracpi^248.endalign$$
Computing $I_44$.
Start the same as with $I_22$ to get $displaystyle I_44=int_0^1 fraclnleft(frac(1+x)^24xright)(1-x)(1-x^4),dx=frac-18int_0^1 fracln(1-t^2)t^2(1+t^2)(1+t)^3dt$.
We can calculate these integrals:
$$beginalign int_0^1 fracln(1-x^2)1+x^2dx=int_0^1 fracln(1+x)1+x^2dx+int_0^1 fracln(1-x)1+x^2dx tag4
\=int_0^1 fracln(1+x)1+x^2dx +int_0^1 fraclnleft(frac2t1+tright)1+t^2dt
\=fracpi4ln2+sum_n=0^infty (-1)^nint_0^1ln(t) t^2ndt=fracpi4ln2-G. endalign$$
$$beginalign int_0^1 fracln(1-x^2)x^2(1+x^2)dx=int_0^1 fracln(1-x^2)x^2dx-int_0^1 fracln(1-x^2)1+x^2dx tag5
\=-sum_n=0^infty frac1n+1int_0^1 x^2ndx-fracpi4ln2+G=G-fracpi4ln2-2ln2.endalign$$
$$beginalign int_0^1 fracxln(1-x^2)1+x^2dx=frac12int_0^1 fracln(1-x)1+xdx tag6
\=-frac12 operatornameLi_2left(frac12right)=fracln^2 24-fracpi^224.endalign$$
$$beginalign int_0^1 fracln(1-x^2)x(1+x^2)dx=int_0^1 fracln(1-x^2)xdx-int_0^1 fracxln(1-x^2)1+x^2dx tag7
\=-sum_n=0^inftyfrac1n+1int_0^1 x^2n+1dx-fracln^2 24+fracpi^224=-fracpi^224-fracln^2 24.endalign$$
Altogether,
$$I_44=frac-18int_0^1 fracln(1-x^2)x^2(1+x^2)(1+3x+3x^2+x^3)dx
\=-fracpi16ln2+fracln24+fracln^2 216+fracpi^248+fracG4.$$
Computing $I_24$.
Substitute $u=fracyx^2$, change the order of integration, let $yto y^2$, evaluate the inner integral,and substitue $t=frac1-x1+x$:
$$beginalign* I_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^2)=int_0^1int_0^x^2 fracdydx(1+x)(x^2+y)(1-y^2)
\=int_0^1 frac11-y^2int_sqrty^1fracdx(1+x)(x^2+y)dy=2int_0^1fracy1-y^4int_y^1fracdx(1+x)(x^2+y^2)dy
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1-x^4)dx
=I_241-I_242. endalign*$$
Evaulation of $I_241$.
Substitute $t=frac1-x1+x$ to get $displaystyle I_241=frac14int_0^1 fractan^-1(t)t(1+t^2)^2(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=tantheta$:
$$beginalign int_0^1 fractan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetacos^2(theta)dtheta=fracpi^264+fracpi16-frac18.tag8endalign$$
$$beginalign int_0^1 fracxtan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetatanthetacos^2thetadtheta=frac12int_0^fracpi4thetasin(2theta)dtheta=frac18.tag9endalign$$
$$beginalign int_0^1 fracx^2tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^2(theta)dtheta=fracpi^264-fracpi16+frac18.tag10endalign$$
$$beginalign int_0^1 fracx^3tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^3(theta)sectheta,dtheta tag11
\=int_0^fracpi4thetatantheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=fracpi8ln2-frac18+int_0^fracpi4lncostheta ,dtheta
\=fracpi8ln2-frac18-int_0^fracpi4ln2 ,dtheta-sum_n=1^infty frac(-1)^nnint_0^fracpi4cos(2ntheta),dtheta
\=-fracpi8ln2-frac18+frac12sum_n=1^inftyfrac(-1)^n+1n^2sin(fracpi n2)=fracG2-fracpi8ln2-frac18.endalign$$
$$beginalign int_0^1 fractan^-1(x)x(1+x^2)^2dx=int_0^fracpi4thetacos^3(theta)csctheta,dtheta tag12
\=int_0^fracpi4thetacottheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=-frac18-fracpi8ln2-int_0^fracpi4lnsintheta ,dtheta
\=-frac18-fracpi8ln2+int_0^fracpi4ln2 ,dtheta+sum_n=1^inftyfrac1nint_0^fracpi4cos(2ntheta),dtheta
\=-frac18+fracpi8ln2+frac12sum_n=1^infty fracsin(fracpi n2)n^2=fracG2+fracpi8ln2-frac18.endalign$$
Altogether,
$$I_241=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx= frac14int_0^1 fractan^-1(x)x(1+x^2)^2(1+4x+6x^2+4x^3+x^4)dx
\=fracpi^232+frac18+fracG4$$
Evaulation of $I_242$.
Substitute $t=frac1-x1+x$ to get
$$ I_242=frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)^2(1-t^2)(1+t)^2dt
\=frac12int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)(1+t^2)^2(1-t^2)dt+frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)(1-t^2)dt
\=frac12int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx-frac14int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx\+frac18int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx-frac14int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx
$$
Calculating these integrals:
$$beginalign int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx=-frac12int_0^1 fraclnleft(frac1-x1+xright)sqrtx(1+x)frac1-x1+xdx tag13
\=-frac12int_0^1fractln tsqrt1-t^2dt=-frac18int_0^1fracln tsqrt1-tdt=-frac18int_0^1 t^-1/2ln(1-t)dt
\=frac14sum_n=0^infty frac1(n+1)(2n+3)=frac12-fracln22.endalign$$
$$beginalign int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1+x)(1-x)x(1+x)dx tag14
\=frac12int_0^1fracln(1+x)xdx-int_0^1fracln(1+x)1+xdx
\=frac12sum_n=0^inftyfrac(-1)^n+1n+1int_0^1 x^n dx-frac12ln^2(1+x)bigg_0^1=fracpi^224-fracln^2 22.endalign$$
$$beginalign int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1-x)xdx-int_0^1fracln(1-x)1+xdx tag15
\=-fracpi^212-left(fracln^2 22-fracpi^212right)=-fracln^2 22.endalign$$
$$
int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx=-2int_0^fracpi4cos^2(theta)(1-tan^2theta)lncostheta,dtheta tag16
\=-2int_0^fracpi4cos(2theta)lncostheta,dtheta=2ln2int_0^fracpi4cos(2theta)dtheta+sum_n=1^infty frac(-1)^nnint_0^fracpi2costheta cos(ntheta)dtheta
\=ln2-fracpi4+sum_n=1^infty frac(-1)^2n2nfrac(-1)^n+1(2n)^2-1=fracln22-fracpi4+frac12.$$
Altogether, $displaystyle I_242=fracpi^2192+fracpi16+fracln^2 216-frac3ln28+frac18$,
leading to $displaystyle I_24=I_241-I_242=frac5pi^2192-fracpi16-fracln^2 216+frac3ln28+fracG4$,
and finally, confirming the conjecture,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=I_22-2I_24+I_44=fracpi8-fracpi16ln2-fracpi^296+frac3ln^2 216-fracG4.$$
I don't know about higher powers. I guess the case $mathcal A_2$ can also be done. If we start the same as with $mathcal B_2$, writing $mathcal A_2=J_22-2J_24+J_44$
we can find that $displaystyle J_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^2u^2)=2I_44-I_22=-fracpi8ln2+fracG2+fracpi^248+fracln^2 28$
$J_44$ can be reduced to $displaystyle =-frac12int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2sqrt2)(x+3-2sqrt2).$
I can get $displaystyle int_0^1 fracln(1-x^2)x(x^4+6x^2+1)dx=-fracpi^212+frac4-3sqrt216operatornameLi_2left(frac2-sqrt24right)+frac4+3sqrt216operatornameLi_2left(frac2+sqrt24right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx=frac1+2sqrt24piln2-fracpi^224-frac14lnleft(frac2+sqrt24right)lnleft(frac2-sqrt24right)
-fracsqrt2+12ImoperatornameLi_2left(frac2+sqrt22+fracisqrt22right)-fracsqrt2-12ImoperatornameLi_2left(frac2-sqrt22+fracisqrt22right)$$
I obtained it by calculating $displaystyle int_0^1 fracln(1+x)x+adx=ln2lnleft(fraca+1a-1right)+operatornameLi_2left(frac21-aright)-operatornameLi_2left(frac11-aright)$,
which together with $(3)$ can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x+adx$, which in turn, through partial fractions, can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x^2+a^2dx$.
Fortunately, things didn't get too ugly as both $3+2sqrt2$ and $3-2sqrt2$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_24$. Starting similarly as with $I_24$,
we have:
$$J_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^4u^2)
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1+x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1+x^4)dx
\=J_241-J_242$$
Through $t=frac1-x1+x$, $J_241$ turns to $displaystyle int_0^1 fractan^-1(x)(1+x^2)(x^4+6x^2+1)(1+x)^4,dx$. I don't have any idea about that yet. Edit 1.
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
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So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $displaystyle H_n-2H_2n+H_4n=int_0^1 fracx^2n-x^4n1+xdx$.
I noticed that $H_n-2H_2n+H_4n=H_4n-H_2n-(H_2n-H_n)=H_4n^--H_2n^-$, where $H_n^-=sum_k=1^n frac(-1)^k+1k$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=frac1-x1+x$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -lnsin x=ln2+sum_n=1^infty fraccos(2nx)n ,-lncos x=ln2+sum_n=1^infty frac(-1)^ncos(2nx)n tag1$
$$ int_0^fracpi2 cos x cos(nx)dx=begincases fracpi4 &n=1\0 &n ,,textodd\ frac(-1)^1+n/2n^2-1 &n ,,texteven endcases tag2$$
$$ int_0^1 fracln(1-x)a+xdx=-operatornameLi_2left(frac1a+1right)tag3$$
Starting,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=sum_n=0^inftyint_0^1int_0^1frac(x^2n-x^4n)(u^2n-u^4n)(1+x)(1+u)dxdu
\=smallint_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)-2int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^4)+int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^4)
\=I_22-2I_24+I_44$$
Computing $I_22$.
Substitute $u=fracyx$ ,change the order of integration, evaluate the inner integral, and substitue $t=frac1-x1+x$ to get
$$beginalign I_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)=int_0^1int_0^xfracdydx(1+x)(x+y)(1-y^2)
\=int_0^1 frac11-y^2int_1^y fracdx(1+x)(x+y) dy=int_0^1 fraclnleft(frac(1+x)^24xright)(1+x)(1-x^2),dx
\=frac-14int_0^1 frac(1+t)t^2ln(1-t^2)dt=-frac14int_0^1fracln(1-t^2)t^2dt-frac14int_0^1fracln(1-t^2)tdt
\=frac14sum_n=0^infty frac1(n+1)(2n+1)+frac14sum_n=0^infty frac1(n+1)(2n+2)=fracln22+fracpi^248.endalign$$
Computing $I_44$.
Start the same as with $I_22$ to get $displaystyle I_44=int_0^1 fraclnleft(frac(1+x)^24xright)(1-x)(1-x^4),dx=frac-18int_0^1 fracln(1-t^2)t^2(1+t^2)(1+t)^3dt$.
We can calculate these integrals:
$$beginalign int_0^1 fracln(1-x^2)1+x^2dx=int_0^1 fracln(1+x)1+x^2dx+int_0^1 fracln(1-x)1+x^2dx tag4
\=int_0^1 fracln(1+x)1+x^2dx +int_0^1 fraclnleft(frac2t1+tright)1+t^2dt
\=fracpi4ln2+sum_n=0^infty (-1)^nint_0^1ln(t) t^2ndt=fracpi4ln2-G. endalign$$
$$beginalign int_0^1 fracln(1-x^2)x^2(1+x^2)dx=int_0^1 fracln(1-x^2)x^2dx-int_0^1 fracln(1-x^2)1+x^2dx tag5
\=-sum_n=0^infty frac1n+1int_0^1 x^2ndx-fracpi4ln2+G=G-fracpi4ln2-2ln2.endalign$$
$$beginalign int_0^1 fracxln(1-x^2)1+x^2dx=frac12int_0^1 fracln(1-x)1+xdx tag6
\=-frac12 operatornameLi_2left(frac12right)=fracln^2 24-fracpi^224.endalign$$
$$beginalign int_0^1 fracln(1-x^2)x(1+x^2)dx=int_0^1 fracln(1-x^2)xdx-int_0^1 fracxln(1-x^2)1+x^2dx tag7
\=-sum_n=0^inftyfrac1n+1int_0^1 x^2n+1dx-fracln^2 24+fracpi^224=-fracpi^224-fracln^2 24.endalign$$
Altogether,
$$I_44=frac-18int_0^1 fracln(1-x^2)x^2(1+x^2)(1+3x+3x^2+x^3)dx
\=-fracpi16ln2+fracln24+fracln^2 216+fracpi^248+fracG4.$$
Computing $I_24$.
Substitute $u=fracyx^2$, change the order of integration, let $yto y^2$, evaluate the inner integral,and substitue $t=frac1-x1+x$:
$$beginalign* I_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^2)=int_0^1int_0^x^2 fracdydx(1+x)(x^2+y)(1-y^2)
\=int_0^1 frac11-y^2int_sqrty^1fracdx(1+x)(x^2+y)dy=2int_0^1fracy1-y^4int_y^1fracdx(1+x)(x^2+y^2)dy
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1-x^4)dx
=I_241-I_242. endalign*$$
Evaulation of $I_241$.
Substitute $t=frac1-x1+x$ to get $displaystyle I_241=frac14int_0^1 fractan^-1(t)t(1+t^2)^2(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=tantheta$:
$$beginalign int_0^1 fractan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetacos^2(theta)dtheta=fracpi^264+fracpi16-frac18.tag8endalign$$
$$beginalign int_0^1 fracxtan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetatanthetacos^2thetadtheta=frac12int_0^fracpi4thetasin(2theta)dtheta=frac18.tag9endalign$$
$$beginalign int_0^1 fracx^2tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^2(theta)dtheta=fracpi^264-fracpi16+frac18.tag10endalign$$
$$beginalign int_0^1 fracx^3tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^3(theta)sectheta,dtheta tag11
\=int_0^fracpi4thetatantheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=fracpi8ln2-frac18+int_0^fracpi4lncostheta ,dtheta
\=fracpi8ln2-frac18-int_0^fracpi4ln2 ,dtheta-sum_n=1^infty frac(-1)^nnint_0^fracpi4cos(2ntheta),dtheta
\=-fracpi8ln2-frac18+frac12sum_n=1^inftyfrac(-1)^n+1n^2sin(fracpi n2)=fracG2-fracpi8ln2-frac18.endalign$$
$$beginalign int_0^1 fractan^-1(x)x(1+x^2)^2dx=int_0^fracpi4thetacos^3(theta)csctheta,dtheta tag12
\=int_0^fracpi4thetacottheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=-frac18-fracpi8ln2-int_0^fracpi4lnsintheta ,dtheta
\=-frac18-fracpi8ln2+int_0^fracpi4ln2 ,dtheta+sum_n=1^inftyfrac1nint_0^fracpi4cos(2ntheta),dtheta
\=-frac18+fracpi8ln2+frac12sum_n=1^infty fracsin(fracpi n2)n^2=fracG2+fracpi8ln2-frac18.endalign$$
Altogether,
$$I_241=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx= frac14int_0^1 fractan^-1(x)x(1+x^2)^2(1+4x+6x^2+4x^3+x^4)dx
\=fracpi^232+frac18+fracG4$$
Evaulation of $I_242$.
Substitute $t=frac1-x1+x$ to get
$$ I_242=frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)^2(1-t^2)(1+t)^2dt
\=frac12int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)(1+t^2)^2(1-t^2)dt+frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)(1-t^2)dt
\=frac12int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx-frac14int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx\+frac18int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx-frac14int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx
$$
Calculating these integrals:
$$beginalign int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx=-frac12int_0^1 fraclnleft(frac1-x1+xright)sqrtx(1+x)frac1-x1+xdx tag13
\=-frac12int_0^1fractln tsqrt1-t^2dt=-frac18int_0^1fracln tsqrt1-tdt=-frac18int_0^1 t^-1/2ln(1-t)dt
\=frac14sum_n=0^infty frac1(n+1)(2n+3)=frac12-fracln22.endalign$$
$$beginalign int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1+x)(1-x)x(1+x)dx tag14
\=frac12int_0^1fracln(1+x)xdx-int_0^1fracln(1+x)1+xdx
\=frac12sum_n=0^inftyfrac(-1)^n+1n+1int_0^1 x^n dx-frac12ln^2(1+x)bigg_0^1=fracpi^224-fracln^2 22.endalign$$
$$beginalign int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1-x)xdx-int_0^1fracln(1-x)1+xdx tag15
\=-fracpi^212-left(fracln^2 22-fracpi^212right)=-fracln^2 22.endalign$$
$$
int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx=-2int_0^fracpi4cos^2(theta)(1-tan^2theta)lncostheta,dtheta tag16
\=-2int_0^fracpi4cos(2theta)lncostheta,dtheta=2ln2int_0^fracpi4cos(2theta)dtheta+sum_n=1^infty frac(-1)^nnint_0^fracpi2costheta cos(ntheta)dtheta
\=ln2-fracpi4+sum_n=1^infty frac(-1)^2n2nfrac(-1)^n+1(2n)^2-1=fracln22-fracpi4+frac12.$$
Altogether, $displaystyle I_242=fracpi^2192+fracpi16+fracln^2 216-frac3ln28+frac18$,
leading to $displaystyle I_24=I_241-I_242=frac5pi^2192-fracpi16-fracln^2 216+frac3ln28+fracG4$,
and finally, confirming the conjecture,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=I_22-2I_24+I_44=fracpi8-fracpi16ln2-fracpi^296+frac3ln^2 216-fracG4.$$
I don't know about higher powers. I guess the case $mathcal A_2$ can also be done. If we start the same as with $mathcal B_2$, writing $mathcal A_2=J_22-2J_24+J_44$
we can find that $displaystyle J_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^2u^2)=2I_44-I_22=-fracpi8ln2+fracG2+fracpi^248+fracln^2 28$
$J_44$ can be reduced to $displaystyle =-frac12int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2sqrt2)(x+3-2sqrt2).$
I can get $displaystyle int_0^1 fracln(1-x^2)x(x^4+6x^2+1)dx=-fracpi^212+frac4-3sqrt216operatornameLi_2left(frac2-sqrt24right)+frac4+3sqrt216operatornameLi_2left(frac2+sqrt24right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx=frac1+2sqrt24piln2-fracpi^224-frac14lnleft(frac2+sqrt24right)lnleft(frac2-sqrt24right)
-fracsqrt2+12ImoperatornameLi_2left(frac2+sqrt22+fracisqrt22right)-fracsqrt2-12ImoperatornameLi_2left(frac2-sqrt22+fracisqrt22right)$$
I obtained it by calculating $displaystyle int_0^1 fracln(1+x)x+adx=ln2lnleft(fraca+1a-1right)+operatornameLi_2left(frac21-aright)-operatornameLi_2left(frac11-aright)$,
which together with $(3)$ can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x+adx$, which in turn, through partial fractions, can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x^2+a^2dx$.
Fortunately, things didn't get too ugly as both $3+2sqrt2$ and $3-2sqrt2$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_24$. Starting similarly as with $I_24$,
we have:
$$J_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^4u^2)
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1+x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1+x^4)dx
\=J_241-J_242$$
Through $t=frac1-x1+x$, $J_241$ turns to $displaystyle int_0^1 fractan^-1(x)(1+x^2)(x^4+6x^2+1)(1+x)^4,dx$. I don't have any idea about that yet. Edit 1.
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
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add a comment |
$begingroup$
So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $displaystyle H_n-2H_2n+H_4n=int_0^1 fracx^2n-x^4n1+xdx$.
I noticed that $H_n-2H_2n+H_4n=H_4n-H_2n-(H_2n-H_n)=H_4n^--H_2n^-$, where $H_n^-=sum_k=1^n frac(-1)^k+1k$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=frac1-x1+x$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -lnsin x=ln2+sum_n=1^infty fraccos(2nx)n ,-lncos x=ln2+sum_n=1^infty frac(-1)^ncos(2nx)n tag1$
$$ int_0^fracpi2 cos x cos(nx)dx=begincases fracpi4 &n=1\0 &n ,,textodd\ frac(-1)^1+n/2n^2-1 &n ,,texteven endcases tag2$$
$$ int_0^1 fracln(1-x)a+xdx=-operatornameLi_2left(frac1a+1right)tag3$$
Starting,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=sum_n=0^inftyint_0^1int_0^1frac(x^2n-x^4n)(u^2n-u^4n)(1+x)(1+u)dxdu
\=smallint_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)-2int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^4)+int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^4)
\=I_22-2I_24+I_44$$
Computing $I_22$.
Substitute $u=fracyx$ ,change the order of integration, evaluate the inner integral, and substitue $t=frac1-x1+x$ to get
$$beginalign I_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)=int_0^1int_0^xfracdydx(1+x)(x+y)(1-y^2)
\=int_0^1 frac11-y^2int_1^y fracdx(1+x)(x+y) dy=int_0^1 fraclnleft(frac(1+x)^24xright)(1+x)(1-x^2),dx
\=frac-14int_0^1 frac(1+t)t^2ln(1-t^2)dt=-frac14int_0^1fracln(1-t^2)t^2dt-frac14int_0^1fracln(1-t^2)tdt
\=frac14sum_n=0^infty frac1(n+1)(2n+1)+frac14sum_n=0^infty frac1(n+1)(2n+2)=fracln22+fracpi^248.endalign$$
Computing $I_44$.
Start the same as with $I_22$ to get $displaystyle I_44=int_0^1 fraclnleft(frac(1+x)^24xright)(1-x)(1-x^4),dx=frac-18int_0^1 fracln(1-t^2)t^2(1+t^2)(1+t)^3dt$.
We can calculate these integrals:
$$beginalign int_0^1 fracln(1-x^2)1+x^2dx=int_0^1 fracln(1+x)1+x^2dx+int_0^1 fracln(1-x)1+x^2dx tag4
\=int_0^1 fracln(1+x)1+x^2dx +int_0^1 fraclnleft(frac2t1+tright)1+t^2dt
\=fracpi4ln2+sum_n=0^infty (-1)^nint_0^1ln(t) t^2ndt=fracpi4ln2-G. endalign$$
$$beginalign int_0^1 fracln(1-x^2)x^2(1+x^2)dx=int_0^1 fracln(1-x^2)x^2dx-int_0^1 fracln(1-x^2)1+x^2dx tag5
\=-sum_n=0^infty frac1n+1int_0^1 x^2ndx-fracpi4ln2+G=G-fracpi4ln2-2ln2.endalign$$
$$beginalign int_0^1 fracxln(1-x^2)1+x^2dx=frac12int_0^1 fracln(1-x)1+xdx tag6
\=-frac12 operatornameLi_2left(frac12right)=fracln^2 24-fracpi^224.endalign$$
$$beginalign int_0^1 fracln(1-x^2)x(1+x^2)dx=int_0^1 fracln(1-x^2)xdx-int_0^1 fracxln(1-x^2)1+x^2dx tag7
\=-sum_n=0^inftyfrac1n+1int_0^1 x^2n+1dx-fracln^2 24+fracpi^224=-fracpi^224-fracln^2 24.endalign$$
Altogether,
$$I_44=frac-18int_0^1 fracln(1-x^2)x^2(1+x^2)(1+3x+3x^2+x^3)dx
\=-fracpi16ln2+fracln24+fracln^2 216+fracpi^248+fracG4.$$
Computing $I_24$.
Substitute $u=fracyx^2$, change the order of integration, let $yto y^2$, evaluate the inner integral,and substitue $t=frac1-x1+x$:
$$beginalign* I_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^2)=int_0^1int_0^x^2 fracdydx(1+x)(x^2+y)(1-y^2)
\=int_0^1 frac11-y^2int_sqrty^1fracdx(1+x)(x^2+y)dy=2int_0^1fracy1-y^4int_y^1fracdx(1+x)(x^2+y^2)dy
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1-x^4)dx
=I_241-I_242. endalign*$$
Evaulation of $I_241$.
Substitute $t=frac1-x1+x$ to get $displaystyle I_241=frac14int_0^1 fractan^-1(t)t(1+t^2)^2(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=tantheta$:
$$beginalign int_0^1 fractan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetacos^2(theta)dtheta=fracpi^264+fracpi16-frac18.tag8endalign$$
$$beginalign int_0^1 fracxtan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetatanthetacos^2thetadtheta=frac12int_0^fracpi4thetasin(2theta)dtheta=frac18.tag9endalign$$
$$beginalign int_0^1 fracx^2tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^2(theta)dtheta=fracpi^264-fracpi16+frac18.tag10endalign$$
$$beginalign int_0^1 fracx^3tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^3(theta)sectheta,dtheta tag11
\=int_0^fracpi4thetatantheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=fracpi8ln2-frac18+int_0^fracpi4lncostheta ,dtheta
\=fracpi8ln2-frac18-int_0^fracpi4ln2 ,dtheta-sum_n=1^infty frac(-1)^nnint_0^fracpi4cos(2ntheta),dtheta
\=-fracpi8ln2-frac18+frac12sum_n=1^inftyfrac(-1)^n+1n^2sin(fracpi n2)=fracG2-fracpi8ln2-frac18.endalign$$
$$beginalign int_0^1 fractan^-1(x)x(1+x^2)^2dx=int_0^fracpi4thetacos^3(theta)csctheta,dtheta tag12
\=int_0^fracpi4thetacottheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=-frac18-fracpi8ln2-int_0^fracpi4lnsintheta ,dtheta
\=-frac18-fracpi8ln2+int_0^fracpi4ln2 ,dtheta+sum_n=1^inftyfrac1nint_0^fracpi4cos(2ntheta),dtheta
\=-frac18+fracpi8ln2+frac12sum_n=1^infty fracsin(fracpi n2)n^2=fracG2+fracpi8ln2-frac18.endalign$$
Altogether,
$$I_241=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx= frac14int_0^1 fractan^-1(x)x(1+x^2)^2(1+4x+6x^2+4x^3+x^4)dx
\=fracpi^232+frac18+fracG4$$
Evaulation of $I_242$.
Substitute $t=frac1-x1+x$ to get
$$ I_242=frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)^2(1-t^2)(1+t)^2dt
\=frac12int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)(1+t^2)^2(1-t^2)dt+frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)(1-t^2)dt
\=frac12int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx-frac14int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx\+frac18int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx-frac14int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx
$$
Calculating these integrals:
$$beginalign int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx=-frac12int_0^1 fraclnleft(frac1-x1+xright)sqrtx(1+x)frac1-x1+xdx tag13
\=-frac12int_0^1fractln tsqrt1-t^2dt=-frac18int_0^1fracln tsqrt1-tdt=-frac18int_0^1 t^-1/2ln(1-t)dt
\=frac14sum_n=0^infty frac1(n+1)(2n+3)=frac12-fracln22.endalign$$
$$beginalign int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1+x)(1-x)x(1+x)dx tag14
\=frac12int_0^1fracln(1+x)xdx-int_0^1fracln(1+x)1+xdx
\=frac12sum_n=0^inftyfrac(-1)^n+1n+1int_0^1 x^n dx-frac12ln^2(1+x)bigg_0^1=fracpi^224-fracln^2 22.endalign$$
$$beginalign int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1-x)xdx-int_0^1fracln(1-x)1+xdx tag15
\=-fracpi^212-left(fracln^2 22-fracpi^212right)=-fracln^2 22.endalign$$
$$
int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx=-2int_0^fracpi4cos^2(theta)(1-tan^2theta)lncostheta,dtheta tag16
\=-2int_0^fracpi4cos(2theta)lncostheta,dtheta=2ln2int_0^fracpi4cos(2theta)dtheta+sum_n=1^infty frac(-1)^nnint_0^fracpi2costheta cos(ntheta)dtheta
\=ln2-fracpi4+sum_n=1^infty frac(-1)^2n2nfrac(-1)^n+1(2n)^2-1=fracln22-fracpi4+frac12.$$
Altogether, $displaystyle I_242=fracpi^2192+fracpi16+fracln^2 216-frac3ln28+frac18$,
leading to $displaystyle I_24=I_241-I_242=frac5pi^2192-fracpi16-fracln^2 216+frac3ln28+fracG4$,
and finally, confirming the conjecture,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=I_22-2I_24+I_44=fracpi8-fracpi16ln2-fracpi^296+frac3ln^2 216-fracG4.$$
I don't know about higher powers. I guess the case $mathcal A_2$ can also be done. If we start the same as with $mathcal B_2$, writing $mathcal A_2=J_22-2J_24+J_44$
we can find that $displaystyle J_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^2u^2)=2I_44-I_22=-fracpi8ln2+fracG2+fracpi^248+fracln^2 28$
$J_44$ can be reduced to $displaystyle =-frac12int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2sqrt2)(x+3-2sqrt2).$
I can get $displaystyle int_0^1 fracln(1-x^2)x(x^4+6x^2+1)dx=-fracpi^212+frac4-3sqrt216operatornameLi_2left(frac2-sqrt24right)+frac4+3sqrt216operatornameLi_2left(frac2+sqrt24right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx=frac1+2sqrt24piln2-fracpi^224-frac14lnleft(frac2+sqrt24right)lnleft(frac2-sqrt24right)
-fracsqrt2+12ImoperatornameLi_2left(frac2+sqrt22+fracisqrt22right)-fracsqrt2-12ImoperatornameLi_2left(frac2-sqrt22+fracisqrt22right)$$
I obtained it by calculating $displaystyle int_0^1 fracln(1+x)x+adx=ln2lnleft(fraca+1a-1right)+operatornameLi_2left(frac21-aright)-operatornameLi_2left(frac11-aright)$,
which together with $(3)$ can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x+adx$, which in turn, through partial fractions, can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x^2+a^2dx$.
Fortunately, things didn't get too ugly as both $3+2sqrt2$ and $3-2sqrt2$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_24$. Starting similarly as with $I_24$,
we have:
$$J_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^4u^2)
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1+x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1+x^4)dx
\=J_241-J_242$$
Through $t=frac1-x1+x$, $J_241$ turns to $displaystyle int_0^1 fractan^-1(x)(1+x^2)(x^4+6x^2+1)(1+x)^4,dx$. I don't have any idea about that yet. Edit 1.
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
$endgroup$
So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $displaystyle H_n-2H_2n+H_4n=int_0^1 fracx^2n-x^4n1+xdx$.
I noticed that $H_n-2H_2n+H_4n=H_4n-H_2n-(H_2n-H_n)=H_4n^--H_2n^-$, where $H_n^-=sum_k=1^n frac(-1)^k+1k$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=frac1-x1+x$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -lnsin x=ln2+sum_n=1^infty fraccos(2nx)n ,-lncos x=ln2+sum_n=1^infty frac(-1)^ncos(2nx)n tag1$
$$ int_0^fracpi2 cos x cos(nx)dx=begincases fracpi4 &n=1\0 &n ,,textodd\ frac(-1)^1+n/2n^2-1 &n ,,texteven endcases tag2$$
$$ int_0^1 fracln(1-x)a+xdx=-operatornameLi_2left(frac1a+1right)tag3$$
Starting,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=sum_n=0^inftyint_0^1int_0^1frac(x^2n-x^4n)(u^2n-u^4n)(1+x)(1+u)dxdu
\=smallint_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)-2int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^4)+int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^4)
\=I_22-2I_24+I_44$$
Computing $I_22$.
Substitute $u=fracyx$ ,change the order of integration, evaluate the inner integral, and substitue $t=frac1-x1+x$ to get
$$beginalign I_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^2u^2)=int_0^1int_0^xfracdydx(1+x)(x+y)(1-y^2)
\=int_0^1 frac11-y^2int_1^y fracdx(1+x)(x+y) dy=int_0^1 fraclnleft(frac(1+x)^24xright)(1+x)(1-x^2),dx
\=frac-14int_0^1 frac(1+t)t^2ln(1-t^2)dt=-frac14int_0^1fracln(1-t^2)t^2dt-frac14int_0^1fracln(1-t^2)tdt
\=frac14sum_n=0^infty frac1(n+1)(2n+1)+frac14sum_n=0^infty frac1(n+1)(2n+2)=fracln22+fracpi^248.endalign$$
Computing $I_44$.
Start the same as with $I_22$ to get $displaystyle I_44=int_0^1 fraclnleft(frac(1+x)^24xright)(1-x)(1-x^4),dx=frac-18int_0^1 fracln(1-t^2)t^2(1+t^2)(1+t)^3dt$.
We can calculate these integrals:
$$beginalign int_0^1 fracln(1-x^2)1+x^2dx=int_0^1 fracln(1+x)1+x^2dx+int_0^1 fracln(1-x)1+x^2dx tag4
\=int_0^1 fracln(1+x)1+x^2dx +int_0^1 fraclnleft(frac2t1+tright)1+t^2dt
\=fracpi4ln2+sum_n=0^infty (-1)^nint_0^1ln(t) t^2ndt=fracpi4ln2-G. endalign$$
$$beginalign int_0^1 fracln(1-x^2)x^2(1+x^2)dx=int_0^1 fracln(1-x^2)x^2dx-int_0^1 fracln(1-x^2)1+x^2dx tag5
\=-sum_n=0^infty frac1n+1int_0^1 x^2ndx-fracpi4ln2+G=G-fracpi4ln2-2ln2.endalign$$
$$beginalign int_0^1 fracxln(1-x^2)1+x^2dx=frac12int_0^1 fracln(1-x)1+xdx tag6
\=-frac12 operatornameLi_2left(frac12right)=fracln^2 24-fracpi^224.endalign$$
$$beginalign int_0^1 fracln(1-x^2)x(1+x^2)dx=int_0^1 fracln(1-x^2)xdx-int_0^1 fracxln(1-x^2)1+x^2dx tag7
\=-sum_n=0^inftyfrac1n+1int_0^1 x^2n+1dx-fracln^2 24+fracpi^224=-fracpi^224-fracln^2 24.endalign$$
Altogether,
$$I_44=frac-18int_0^1 fracln(1-x^2)x^2(1+x^2)(1+3x+3x^2+x^3)dx
\=-fracpi16ln2+fracln24+fracln^2 216+fracpi^248+fracG4.$$
Computing $I_24$.
Substitute $u=fracyx^2$, change the order of integration, let $yto y^2$, evaluate the inner integral,and substitue $t=frac1-x1+x$:
$$beginalign* I_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1-x^4u^2)=int_0^1int_0^x^2 fracdydx(1+x)(x^2+y)(1-y^2)
\=int_0^1 frac11-y^2int_sqrty^1fracdx(1+x)(x^2+y)dy=2int_0^1fracy1-y^4int_y^1fracdx(1+x)(x^2+y^2)dy
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1-x^4)dx
=I_241-I_242. endalign*$$
Evaulation of $I_241$.
Substitute $t=frac1-x1+x$ to get $displaystyle I_241=frac14int_0^1 fractan^-1(t)t(1+t^2)^2(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=tantheta$:
$$beginalign int_0^1 fractan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetacos^2(theta)dtheta=fracpi^264+fracpi16-frac18.tag8endalign$$
$$beginalign int_0^1 fracxtan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetatanthetacos^2thetadtheta=frac12int_0^fracpi4thetasin(2theta)dtheta=frac18.tag9endalign$$
$$beginalign int_0^1 fracx^2tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^2(theta)dtheta=fracpi^264-fracpi16+frac18.tag10endalign$$
$$beginalign int_0^1 fracx^3tan^-1(x)(1+x^2)^2dx=int_0^fracpi4thetasin^3(theta)sectheta,dtheta tag11
\=int_0^fracpi4thetatantheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=fracpi8ln2-frac18+int_0^fracpi4lncostheta ,dtheta
\=fracpi8ln2-frac18-int_0^fracpi4ln2 ,dtheta-sum_n=1^infty frac(-1)^nnint_0^fracpi4cos(2ntheta),dtheta
\=-fracpi8ln2-frac18+frac12sum_n=1^inftyfrac(-1)^n+1n^2sin(fracpi n2)=fracG2-fracpi8ln2-frac18.endalign$$
$$beginalign int_0^1 fractan^-1(x)x(1+x^2)^2dx=int_0^fracpi4thetacos^3(theta)csctheta,dtheta tag12
\=int_0^fracpi4thetacottheta ,dtheta-int_0^fracpi4thetasinthetacostheta ,dtheta=-frac18-fracpi8ln2-int_0^fracpi4lnsintheta ,dtheta
\=-frac18-fracpi8ln2+int_0^fracpi4ln2 ,dtheta+sum_n=1^inftyfrac1nint_0^fracpi4cos(2ntheta),dtheta
\=-frac18+fracpi8ln2+frac12sum_n=1^infty fracsin(fracpi n2)n^2=fracG2+fracpi8ln2-frac18.endalign$$
Altogether,
$$I_241=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1-x^4)dx= frac14int_0^1 fractan^-1(x)x(1+x^2)^2(1+4x+6x^2+4x^3+x^4)dx
\=fracpi^232+frac18+fracG4$$
Evaulation of $I_242$.
Substitute $t=frac1-x1+x$ to get
$$ I_242=frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)^2(1-t^2)(1+t)^2dt
\=frac12int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)(1+t^2)^2(1-t^2)dt+frac14int_0^1 fraclnleft(fracsqrt1+t^21-t^2right)t(1+t^2)(1-t^2)dt
\=frac12int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx-frac14int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx\+frac18int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx-frac14int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx
$$
Calculating these integrals:
$$beginalign int_0^1 fraclnleft(frac1+x^21-x^2right)(1+x^2)^2(1-x^2)dx=-frac12int_0^1 fraclnleft(frac1-x1+xright)sqrtx(1+x)frac1-x1+xdx tag13
\=-frac12int_0^1fractln tsqrt1-t^2dt=-frac18int_0^1fracln tsqrt1-tdt=-frac18int_0^1 t^-1/2ln(1-t)dt
\=frac14sum_n=0^infty frac1(n+1)(2n+3)=frac12-fracln22.endalign$$
$$beginalign int_0^1fracln(1+x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1+x)(1-x)x(1+x)dx tag14
\=frac12int_0^1fracln(1+x)xdx-int_0^1fracln(1+x)1+xdx
\=frac12sum_n=0^inftyfrac(-1)^n+1n+1int_0^1 x^n dx-frac12ln^2(1+x)bigg_0^1=fracpi^224-fracln^2 22.endalign$$
$$beginalign int_0^1fracln(1-x^2)(1-x^2)x(1+x^2)dx=frac12int_0^1fracln(1-x)xdx-int_0^1fracln(1-x)1+xdx tag15
\=-fracpi^212-left(fracln^2 22-fracpi^212right)=-fracln^2 22.endalign$$
$$
int_0^1fracln(1+x^2)(1+x^2)^2(1-x^2)dx=-2int_0^fracpi4cos^2(theta)(1-tan^2theta)lncostheta,dtheta tag16
\=-2int_0^fracpi4cos(2theta)lncostheta,dtheta=2ln2int_0^fracpi4cos(2theta)dtheta+sum_n=1^infty frac(-1)^nnint_0^fracpi2costheta cos(ntheta)dtheta
\=ln2-fracpi4+sum_n=1^infty frac(-1)^2n2nfrac(-1)^n+1(2n)^2-1=fracln22-fracpi4+frac12.$$
Altogether, $displaystyle I_242=fracpi^2192+fracpi16+fracln^2 216-frac3ln28+frac18$,
leading to $displaystyle I_24=I_241-I_242=frac5pi^2192-fracpi16-fracln^2 216+frac3ln28+fracG4$,
and finally, confirming the conjecture,
$$sum_n=0^infty(H_n-2H_2n+H_4n)^2=I_22-2I_24+I_44=fracpi8-fracpi16ln2-fracpi^296+frac3ln^2 216-fracG4.$$
I don't know about higher powers. I guess the case $mathcal A_2$ can also be done. If we start the same as with $mathcal B_2$, writing $mathcal A_2=J_22-2J_24+J_44$
we can find that $displaystyle J_22=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^2u^2)=2I_44-I_22=-fracpi8ln2+fracG2+fracpi^248+fracln^2 28$
$J_44$ can be reduced to $displaystyle =-frac12int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2sqrt2)(x+3-2sqrt2).$
I can get $displaystyle int_0^1 fracln(1-x^2)x(x^4+6x^2+1)dx=-fracpi^212+frac4-3sqrt216operatornameLi_2left(frac2-sqrt24right)+frac4+3sqrt216operatornameLi_2left(frac2+sqrt24right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$int_0^1 fracln(1-x^2)x(x^4+6x^2+1)(1+x)^3 dx=frac1+2sqrt24piln2-fracpi^224-frac14lnleft(frac2+sqrt24right)lnleft(frac2-sqrt24right)
-fracsqrt2+12ImoperatornameLi_2left(frac2+sqrt22+fracisqrt22right)-fracsqrt2-12ImoperatornameLi_2left(frac2-sqrt22+fracisqrt22right)$$
I obtained it by calculating $displaystyle int_0^1 fracln(1+x)x+adx=ln2lnleft(fraca+1a-1right)+operatornameLi_2left(frac21-aright)-operatornameLi_2left(frac11-aright)$,
which together with $(3)$ can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x+adx$, which in turn, through partial fractions, can be used to give a closed form for $displaystyle int_0^1fracln(1-x^2)x^2+a^2dx$.
Fortunately, things didn't get too ugly as both $3+2sqrt2$ and $3-2sqrt2$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_24$. Starting similarly as with $I_24$,
we have:
$$J_24=int_0^1int_0^1fracdxdu(1+x)(1+u)(1+x^4u^2)
\=2int_0^1fractan^-1left(frac1-x1+xright)(1+x^2)(1+x^4)dx-2int_0^1fracxlnleft(frac(1+x)sqrt1+x^22sqrt2xright)(1+x^2)(1+x^4)dx
\=J_241-J_242$$
Through $t=frac1-x1+x$, $J_241$ turns to $displaystyle int_0^1 fractan^-1(x)(1+x^2)(x^4+6x^2+1)(1+x)^4,dx$. I don't have any idea about that yet. Edit 1.
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
edited Sep 15 '15 at 15:09
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
answered Sep 14 '15 at 15:51
nospoonnospoon
4,6361532
4,6361532
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Just my thoughts for now: I would try to exploit Parseval's identity. For first, we have:
$$ H_n-2H_2n+H_4n =int_0^1frac-x^n+2x^2n-x^4n1-x,dx tag1 $$
and:
$$ sum_ngeq 1frac-x^n+2x^2n-x^4n1-xe^niy = frac11-xleft(frac-11-e^iyx+frac21-e^iyx^2+frac-11-e^iyx^4right).tag2$$
The poles of the RHS (as a function of $x$) are located at $xinlefte^-iy,pm e^-iy/2,pm e^-iy/4,pm i e^-iy/4right$.
By using the residue theorem we may compute an explicit representation for:
$$ g(y) = sum_ngeq 1left(H_n-2H_2n+H_4nright)e^niy,tag3 $$
then Parseval's theorem gives:
$$ sum_ngeq 1left(H_n-2H_2n+H_4nright)^2 = frac12piint_-pi^pig(y)g(-y),dy tag4$$
and the resulting integral should be not to difficult to evaluate in terms of dilogarithms.
Another chance may be to apply summation by parts (like I did in this question), but it looks lengthy.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
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It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
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– Winther
Sep 1 '15 at 15:26
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@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
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– Jack D'Aurizio
Sep 1 '15 at 15:35
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Just my thoughts for now: I would try to exploit Parseval's identity. For first, we have:
$$ H_n-2H_2n+H_4n =int_0^1frac-x^n+2x^2n-x^4n1-x,dx tag1 $$
and:
$$ sum_ngeq 1frac-x^n+2x^2n-x^4n1-xe^niy = frac11-xleft(frac-11-e^iyx+frac21-e^iyx^2+frac-11-e^iyx^4right).tag2$$
The poles of the RHS (as a function of $x$) are located at $xinlefte^-iy,pm e^-iy/2,pm e^-iy/4,pm i e^-iy/4right$.
By using the residue theorem we may compute an explicit representation for:
$$ g(y) = sum_ngeq 1left(H_n-2H_2n+H_4nright)e^niy,tag3 $$
then Parseval's theorem gives:
$$ sum_ngeq 1left(H_n-2H_2n+H_4nright)^2 = frac12piint_-pi^pig(y)g(-y),dy tag4$$
and the resulting integral should be not to difficult to evaluate in terms of dilogarithms.
Another chance may be to apply summation by parts (like I did in this question), but it looks lengthy.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
$endgroup$
– Winther
Sep 1 '15 at 15:26
$begingroup$
@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
$endgroup$
– Jack D'Aurizio
Sep 1 '15 at 15:35
add a comment |
$begingroup$
Just my thoughts for now: I would try to exploit Parseval's identity. For first, we have:
$$ H_n-2H_2n+H_4n =int_0^1frac-x^n+2x^2n-x^4n1-x,dx tag1 $$
and:
$$ sum_ngeq 1frac-x^n+2x^2n-x^4n1-xe^niy = frac11-xleft(frac-11-e^iyx+frac21-e^iyx^2+frac-11-e^iyx^4right).tag2$$
The poles of the RHS (as a function of $x$) are located at $xinlefte^-iy,pm e^-iy/2,pm e^-iy/4,pm i e^-iy/4right$.
By using the residue theorem we may compute an explicit representation for:
$$ g(y) = sum_ngeq 1left(H_n-2H_2n+H_4nright)e^niy,tag3 $$
then Parseval's theorem gives:
$$ sum_ngeq 1left(H_n-2H_2n+H_4nright)^2 = frac12piint_-pi^pig(y)g(-y),dy tag4$$
and the resulting integral should be not to difficult to evaluate in terms of dilogarithms.
Another chance may be to apply summation by parts (like I did in this question), but it looks lengthy.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
Just my thoughts for now: I would try to exploit Parseval's identity. For first, we have:
$$ H_n-2H_2n+H_4n =int_0^1frac-x^n+2x^2n-x^4n1-x,dx tag1 $$
and:
$$ sum_ngeq 1frac-x^n+2x^2n-x^4n1-xe^niy = frac11-xleft(frac-11-e^iyx+frac21-e^iyx^2+frac-11-e^iyx^4right).tag2$$
The poles of the RHS (as a function of $x$) are located at $xinlefte^-iy,pm e^-iy/2,pm e^-iy/4,pm i e^-iy/4right$.
By using the residue theorem we may compute an explicit representation for:
$$ g(y) = sum_ngeq 1left(H_n-2H_2n+H_4nright)e^niy,tag3 $$
then Parseval's theorem gives:
$$ sum_ngeq 1left(H_n-2H_2n+H_4nright)^2 = frac12piint_-pi^pig(y)g(-y),dy tag4$$
and the resulting integral should be not to difficult to evaluate in terms of dilogarithms.
Another chance may be to apply summation by parts (like I did in this question), but it looks lengthy.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
answered Aug 30 '15 at 17:36
Jack D'AurizioJack D'Aurizio
291k33284667
291k33284667
$begingroup$
It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
$endgroup$
– Winther
Sep 1 '15 at 15:26
$begingroup$
@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
$endgroup$
– Jack D'Aurizio
Sep 1 '15 at 15:35
add a comment |
$begingroup$
It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
$endgroup$
– Winther
Sep 1 '15 at 15:26
$begingroup$
@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
$endgroup$
– Jack D'Aurizio
Sep 1 '15 at 15:35
$begingroup$
It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
$endgroup$
– Winther
Sep 1 '15 at 15:26
$begingroup$
It’s a very nice idea, but I think you are underestimating the difficulty of evaluating the integral. I checked the representation for $g$ using software: the result is a ton of terms of rational functions of trig’s and inverse trig’s and that is before computing $g(y)g(-y)$. This approach might work, but it looks like a lot of work.
$endgroup$
– Winther
Sep 1 '15 at 15:26
$begingroup$
@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
$endgroup$
– Jack D'Aurizio
Sep 1 '15 at 15:35
$begingroup$
@Winther: I know, it is a tough nut to crack with bare hands, but with some human-guided simplifications, it just boils down to computing some dilogarithmic integrals related with the fourth roots of unity. I haven't really delved into summation by parts, yet. It looks promising, maybe it is the simple way.
$endgroup$
– Jack D'Aurizio
Sep 1 '15 at 15:35
add a comment |
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$begingroup$
Some possibly related questions are linked from here.
$endgroup$
– Piotr Shatalin
Aug 22 '15 at 18:18
3
$begingroup$
BTW, another similar series can be evaluated using generating functions: $sum_n=1^inftyfrac1nleft(H_n-,2H_2n+H_4nright)=frac34 ln^2 2-fracpi^248.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:07
2
$begingroup$
And another: $sum_n=1^inftyfrac(-1)^nnleft(H_n-,2H_2n+H_4nright)=frac12 ln^2!left(1+sqrt2right)+frac18 ln^2 2-frac5pi^296.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 19:43
$begingroup$
One way to tackle it, at least the case $m=2$, would be to write $H_n=int_0^1frac1-x^n1-xdx$ and to convert the series into a double integral which probably could be evaluated more easily. But I think this is what we call "Brute Force".
$endgroup$
– Redundant Aunt
Aug 22 '15 at 19:56
4
$begingroup$
One more simple series where $O(n^-1)$ term is cancelled using harmonic numbers only: $sum_n=1^inftyleft(H_n-4H_2n+5H_4n-2H_8nright)=fracpi4sqrt2 - frac3pi16.$
$endgroup$
– Vladimir Reshetnikov
Aug 22 '15 at 22:14