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A tough series related with a hypergeometric function with quarter integer parameters
Closed-form of $int_0^1 fracln^2(x)sqrtx(a-bx),dx$Definite integral of arcsine over square-root of quadraticHow to prove closed form for a binomial sum?Identities related to hypergeometric functionsAbout a sequence related with the complete elliptic integral of the second kindSimplifying real part of hypergeometric function with complex parametersHypergeometric sum with two binomial factors both in the numerator and in the denominator.An elementary proof of $int_0^1fracarctan xsqrtx(1-x^2),dx = frac132sqrt2pi,Gammaleft(tfrac14right)^2$Looking for a closed form for a $_4 F_3left(ldots,1right)$Generating function of the sequence $binom2nn^3H_n$Infinite series with harmonic numbers related to elliptic integralsAbout the integral $int_0^1fracarctan(x),dxsqrtx(1-x^2)$ and related hypergeometric functionsAn integral inequality related with Hypergeometric function
$begingroup$
Is it possible to express $$ sum_ngeq
0fracbinom4n2nbinom2nn64^n(4n+1) = phantom_3
F_2left(frac14,frac14,frac34; 1,frac54; 1right)
$$ in terms of standard mathematical constants given by Euler sums and values of the $Gamma$ function?
This problem arise from studying the interplay between elliptic integrals, hypergeometric functions and Fourier-Legendre expansions. According to Mathematica's notation we have
$$ sum_ngeq 0fracbinom4n2nbinom2nn64^ny^2n=frac2pisqrt1+y,Kleft(frac2y1+yright) $$
for any $yin[0,1)$, where the complete elliptic integral of the first kind fulfills the functional identity
$$forall xin[0,1),qquad K(x) = fracpi2cdottextAGMleft(1,sqrt1-xright) $$
hence the computation of the above series boils down to the computation of
$$ int_0^1Kleft(frac2y^21+y^2right)frac2,dypisqrt1+y^2stackrelymapstosqrtfracx2-x=fracsqrt2piint_0^1fracK(x)sqrtx(2-x),dx\=frac1piint_-1/2^+inftyfracarctansqrtusqrtu(1+u)(1+2u),du$$
where $K(x),sqrt2-x,frac1sqrtx,frac1sqrt2-x$ all have a pretty simple FL expansion, allowing an easy explicit evaluation of similar integrals. This one, however, is a tougher nut to crack, since $frac12-x$ does not have a nice FL expansion. There are good reasons for believing $Gammaleft(frac14right)$ is involved, since a related series fulfills the following identity:
$$ sum_ngeq 0fracbinom2nn^216^n(4n+1)=frac12piint_0^1K(x),x^-3/4,dx = frac116pi^2,Gammaleft(frac14right)^4 $$
which ultimately is a consequence of Clausen's formula, stating that in some particular circumstances the square of a $phantom_2 F_1$ function is a $phantom_3 F_2$ function.
March 2019 Update: after some manipulations, it turns out that the computation of the original $phantom_3 F_2$ is equivalent to the computation of the integral
$$ int_0^1frac-log xsqrtx(1+6x+x^2),dx = frac1sqrt2int_0^+inftyfracz,dzsqrt3+cosh z$$
which is way less scary. Additionally, nospoon has already tackled similar integrals, so I guess he might have something interesting to share.
sequences-and-series special-functions hypergeometric-function elliptic-integrals
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
$begingroup$
Is it possible to express $$ sum_ngeq
0fracbinom4n2nbinom2nn64^n(4n+1) = phantom_3
F_2left(frac14,frac14,frac34; 1,frac54; 1right)
$$ in terms of standard mathematical constants given by Euler sums and values of the $Gamma$ function?
This problem arise from studying the interplay between elliptic integrals, hypergeometric functions and Fourier-Legendre expansions. According to Mathematica's notation we have
$$ sum_ngeq 0fracbinom4n2nbinom2nn64^ny^2n=frac2pisqrt1+y,Kleft(frac2y1+yright) $$
for any $yin[0,1)$, where the complete elliptic integral of the first kind fulfills the functional identity
$$forall xin[0,1),qquad K(x) = fracpi2cdottextAGMleft(1,sqrt1-xright) $$
hence the computation of the above series boils down to the computation of
$$ int_0^1Kleft(frac2y^21+y^2right)frac2,dypisqrt1+y^2stackrelymapstosqrtfracx2-x=fracsqrt2piint_0^1fracK(x)sqrtx(2-x),dx\=frac1piint_-1/2^+inftyfracarctansqrtusqrtu(1+u)(1+2u),du$$
where $K(x),sqrt2-x,frac1sqrtx,frac1sqrt2-x$ all have a pretty simple FL expansion, allowing an easy explicit evaluation of similar integrals. This one, however, is a tougher nut to crack, since $frac12-x$ does not have a nice FL expansion. There are good reasons for believing $Gammaleft(frac14right)$ is involved, since a related series fulfills the following identity:
$$ sum_ngeq 0fracbinom2nn^216^n(4n+1)=frac12piint_0^1K(x),x^-3/4,dx = frac116pi^2,Gammaleft(frac14right)^4 $$
which ultimately is a consequence of Clausen's formula, stating that in some particular circumstances the square of a $phantom_2 F_1$ function is a $phantom_3 F_2$ function.
March 2019 Update: after some manipulations, it turns out that the computation of the original $phantom_3 F_2$ is equivalent to the computation of the integral
$$ int_0^1frac-log xsqrtx(1+6x+x^2),dx = frac1sqrt2int_0^+inftyfracz,dzsqrt3+cosh z$$
which is way less scary. Additionally, nospoon has already tackled similar integrals, so I guess he might have something interesting to share.
sequences-and-series special-functions hypergeometric-function elliptic-integrals
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
1
$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44
add a comment |
$begingroup$
Is it possible to express $$ sum_ngeq
0fracbinom4n2nbinom2nn64^n(4n+1) = phantom_3
F_2left(frac14,frac14,frac34; 1,frac54; 1right)
$$ in terms of standard mathematical constants given by Euler sums and values of the $Gamma$ function?
This problem arise from studying the interplay between elliptic integrals, hypergeometric functions and Fourier-Legendre expansions. According to Mathematica's notation we have
$$ sum_ngeq 0fracbinom4n2nbinom2nn64^ny^2n=frac2pisqrt1+y,Kleft(frac2y1+yright) $$
for any $yin[0,1)$, where the complete elliptic integral of the first kind fulfills the functional identity
$$forall xin[0,1),qquad K(x) = fracpi2cdottextAGMleft(1,sqrt1-xright) $$
hence the computation of the above series boils down to the computation of
$$ int_0^1Kleft(frac2y^21+y^2right)frac2,dypisqrt1+y^2stackrelymapstosqrtfracx2-x=fracsqrt2piint_0^1fracK(x)sqrtx(2-x),dx\=frac1piint_-1/2^+inftyfracarctansqrtusqrtu(1+u)(1+2u),du$$
where $K(x),sqrt2-x,frac1sqrtx,frac1sqrt2-x$ all have a pretty simple FL expansion, allowing an easy explicit evaluation of similar integrals. This one, however, is a tougher nut to crack, since $frac12-x$ does not have a nice FL expansion. There are good reasons for believing $Gammaleft(frac14right)$ is involved, since a related series fulfills the following identity:
$$ sum_ngeq 0fracbinom2nn^216^n(4n+1)=frac12piint_0^1K(x),x^-3/4,dx = frac116pi^2,Gammaleft(frac14right)^4 $$
which ultimately is a consequence of Clausen's formula, stating that in some particular circumstances the square of a $phantom_2 F_1$ function is a $phantom_3 F_2$ function.
March 2019 Update: after some manipulations, it turns out that the computation of the original $phantom_3 F_2$ is equivalent to the computation of the integral
$$ int_0^1frac-log xsqrtx(1+6x+x^2),dx = frac1sqrt2int_0^+inftyfracz,dzsqrt3+cosh z$$
which is way less scary. Additionally, nospoon has already tackled similar integrals, so I guess he might have something interesting to share.
sequences-and-series special-functions hypergeometric-function elliptic-integrals
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
Is it possible to express $$ sum_ngeq
0fracbinom4n2nbinom2nn64^n(4n+1) = phantom_3
F_2left(frac14,frac14,frac34; 1,frac54; 1right)
$$ in terms of standard mathematical constants given by Euler sums and values of the $Gamma$ function?
This problem arise from studying the interplay between elliptic integrals, hypergeometric functions and Fourier-Legendre expansions. According to Mathematica's notation we have
$$ sum_ngeq 0fracbinom4n2nbinom2nn64^ny^2n=frac2pisqrt1+y,Kleft(frac2y1+yright) $$
for any $yin[0,1)$, where the complete elliptic integral of the first kind fulfills the functional identity
$$forall xin[0,1),qquad K(x) = fracpi2cdottextAGMleft(1,sqrt1-xright) $$
hence the computation of the above series boils down to the computation of
$$ int_0^1Kleft(frac2y^21+y^2right)frac2,dypisqrt1+y^2stackrelymapstosqrtfracx2-x=fracsqrt2piint_0^1fracK(x)sqrtx(2-x),dx\=frac1piint_-1/2^+inftyfracarctansqrtusqrtu(1+u)(1+2u),du$$
where $K(x),sqrt2-x,frac1sqrtx,frac1sqrt2-x$ all have a pretty simple FL expansion, allowing an easy explicit evaluation of similar integrals. This one, however, is a tougher nut to crack, since $frac12-x$ does not have a nice FL expansion. There are good reasons for believing $Gammaleft(frac14right)$ is involved, since a related series fulfills the following identity:
$$ sum_ngeq 0fracbinom2nn^216^n(4n+1)=frac12piint_0^1K(x),x^-3/4,dx = frac116pi^2,Gammaleft(frac14right)^4 $$
which ultimately is a consequence of Clausen's formula, stating that in some particular circumstances the square of a $phantom_2 F_1$ function is a $phantom_3 F_2$ function.
March 2019 Update: after some manipulations, it turns out that the computation of the original $phantom_3 F_2$ is equivalent to the computation of the integral
$$ int_0^1frac-log xsqrtx(1+6x+x^2),dx = frac1sqrt2int_0^+inftyfracz,dzsqrt3+cosh z$$
which is way less scary. Additionally, nospoon has already tackled similar integrals, so I guess he might have something interesting to share.
sequences-and-series special-functions hypergeometric-function elliptic-integrals
sequences-and-series special-functions hypergeometric-function elliptic-integrals
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
edited yesterday
Jack D'Aurizio
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
asked Sep 15 '17 at 0:05
Jack D'AurizioJack D'Aurizio
291k33284666
291k33284666
1
$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44
add a comment |
1
$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44
1
1
$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach.
Denote :
beginequation
S(y):= sumlimits_n=0^infty binom4 n2 n binom2 nn fracy^2 n64^n
endequation
We use the good old identity:
beginequation
binom2 nn=(-4)^n cdot binom-frac12n
endequation
and we get
begineqnarray
S(y)&=& sumlimits_n=0^infty binom-frac12n cdot underbracebinom-frac122n_frac1pi intlimits_0^1 t^-1/2 (1-t)^2 n-1/2 dt cdot (-y^2)^n\
&=& frac1pi intlimits_0^1 frac1sqrtt(1-t)(1-y^2 t^2) dt
&=&
endeqnarray
Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives:
begineqnarray
intlimits_0^1 S(y^2) dy&=& frac1pi intlimits_0^1 fracF_2,1[frac14,frac12,frac54,t^2]sqrtt(1-t)dt\
&=&frac1pi intlimits_0^1 fracF[arcsin(sqrtt),-1]t sqrt1-t dt\
&=&-frac1piintlimits_0^1 frac2log(1-sqrt1-t)-log(t)2sqrtt(1-t^2)dt\
&=&-frac1pi left(2intlimits_0^1 fraclog(t)sqrtt(t-2)(t-1+sqrt2)(t-1-sqrt2)dt+pi^3/2 fracGamma(5/4)Gamma(3/4)right)\
&=&-frac1pi left(
2 sqrt2 intlimits_0^pi fraclog[sin(u/4)]sqrt3-cos(u)-
frac2 sqrtpi Gamma(5/4) (pi+4 log(2))Gamma(-1/4)
right)\
&=&-frac1pi left(
sqrt2 intlimits_0^pi fraclog[1-cos(u/2)]sqrt3-cos(u)du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)\
&=&
-frac1pi left(
sqrt2 intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)
endeqnarray
In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-sqrt1-t$.In the fifthe line I substituted for $t:=2 sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result:
begineqnarray
&&intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du =
-fracpi2sumlimits_lambda=0^infty
binom-1/2lambda (-frac12)^lambda cdot \
&&
left(frac1+lambda1+2lambda binom1/2+lambda-1/2(2 log(2)+H_lambda) + binomlambda-1/2 F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1] right)
endeqnarray
Now, the remaining hypergeometric functions $F_3,2$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity:
beginequation
frac(1+lambda)^(l)(3/2+lambda)^(l) = frac(3/2)^(lambda)1^(lambda) cdot frac1^(l)(3/2)^(l) cdot frac(l+1)^(lambda)(l+3/2)^(lambda)
endequation
and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity:
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C- sumlimits_j=1^lambda j binom-1/2+jlambda binomlambdaj (-1)^lambda-j cdot intlimits_0^1 theta^j-1/2 cdot F_3,2[beginarrayrrr frac12& 1&1\ frac32& frac32 & endarray;theta]dthetaright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^l j binom-1/2j binom-1/2l-j cdot right.\
&&left.intlimits_0^pi/2 [sin(u)]^2j-1 2 cos(u)left( i textLi_2left(-e^i uright)-i textLi_2left(e^i uright)+u log left(frac1-e^i u1+e^i uright)right)duright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^lsumlimits_p=-j-1^j-1 j binom-1/2j binom-1/2l-j cdot frac(-1)^p+l2^2j-1[binom2j-1p+1 - binom2j-1p+j+1]cdotright.\
&&left.intlimits_1^imath z^p+1 left(log(z) log(frac1-z1+z)+Li_2(z)-Li_2(-z)right)dzright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C+right.\
&&left.
frac12
sumlimits_j=0^l-1sumlimits_p=-l-1,pneq -1^l-1
binom-3/2j binom-1/2l-1-j frac12^2j+1
[binom2j+1p+j+1-binom2j+1p+j+2]cdotright.\
&&left.
frac(-1)^p+lp+1
left((1+(-1)^p) cdot C - sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2right)right.\
&&left.
right)
endeqnarray
where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=exp(imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.
In summary we have shown that :
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
2 frac(1/2)^(lambda) (3/2)^(lambda)lambda! cdot lambda! cdot C + frac(3/2)^(lambda)lambda!cdot mathfrak A_lambda\
&&intlimits_0^pi/2 fraclog(1-cos(u))sqrt1-1/2 cos(u)^2=fracsqrtfrac2pi Gamma left(frac14right) (4 C+pi log (2))Gamma left(-frac14right)-\
&& sumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)\
&&intlimits_0^1 S(y^2) dy=-frac1pileft(right.\
&&left.
-frac2 Gamma left(frac54right) (pi (pi -2 log (4))-16 C)sqrtpi Gamma left(-frac14right)right.\
&&left.
-sqrt2cdotsumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)right.\
&&left.
right)
endeqnarray
where
begineqnarray
&&mathfrak A_lambda := \
&&left(
sumlimits_j=0^lambda-1 sumlimits_beginarrayr p=-lambda-1\pneq-1endarray^lambda-1
binom-3/2j binom-1/2lambda-1-j frac[binom2j+1p+j+1 - binom2j+1p+j+2]2^2 j+2
frac(-1)^p+lambda+1p+1cdot
sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2
right)
endeqnarray
Note : The remaining infinite sum converges quite quickly. Truncating the sum at the first one hundred terms produces a more than thirty digits accuracy. On the other hand the the series in question converges very slowly; one needs to take five thousand terms in order to get the first four decimal digits right.
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
$endgroup$
add a comment |
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$begingroup$
This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach.
Denote :
beginequation
S(y):= sumlimits_n=0^infty binom4 n2 n binom2 nn fracy^2 n64^n
endequation
We use the good old identity:
beginequation
binom2 nn=(-4)^n cdot binom-frac12n
endequation
and we get
begineqnarray
S(y)&=& sumlimits_n=0^infty binom-frac12n cdot underbracebinom-frac122n_frac1pi intlimits_0^1 t^-1/2 (1-t)^2 n-1/2 dt cdot (-y^2)^n\
&=& frac1pi intlimits_0^1 frac1sqrtt(1-t)(1-y^2 t^2) dt
&=&
endeqnarray
Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives:
begineqnarray
intlimits_0^1 S(y^2) dy&=& frac1pi intlimits_0^1 fracF_2,1[frac14,frac12,frac54,t^2]sqrtt(1-t)dt\
&=&frac1pi intlimits_0^1 fracF[arcsin(sqrtt),-1]t sqrt1-t dt\
&=&-frac1piintlimits_0^1 frac2log(1-sqrt1-t)-log(t)2sqrtt(1-t^2)dt\
&=&-frac1pi left(2intlimits_0^1 fraclog(t)sqrtt(t-2)(t-1+sqrt2)(t-1-sqrt2)dt+pi^3/2 fracGamma(5/4)Gamma(3/4)right)\
&=&-frac1pi left(
2 sqrt2 intlimits_0^pi fraclog[sin(u/4)]sqrt3-cos(u)-
frac2 sqrtpi Gamma(5/4) (pi+4 log(2))Gamma(-1/4)
right)\
&=&-frac1pi left(
sqrt2 intlimits_0^pi fraclog[1-cos(u/2)]sqrt3-cos(u)du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)\
&=&
-frac1pi left(
sqrt2 intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)
endeqnarray
In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-sqrt1-t$.In the fifthe line I substituted for $t:=2 sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result:
begineqnarray
&&intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du =
-fracpi2sumlimits_lambda=0^infty
binom-1/2lambda (-frac12)^lambda cdot \
&&
left(frac1+lambda1+2lambda binom1/2+lambda-1/2(2 log(2)+H_lambda) + binomlambda-1/2 F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1] right)
endeqnarray
Now, the remaining hypergeometric functions $F_3,2$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity:
beginequation
frac(1+lambda)^(l)(3/2+lambda)^(l) = frac(3/2)^(lambda)1^(lambda) cdot frac1^(l)(3/2)^(l) cdot frac(l+1)^(lambda)(l+3/2)^(lambda)
endequation
and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity:
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C- sumlimits_j=1^lambda j binom-1/2+jlambda binomlambdaj (-1)^lambda-j cdot intlimits_0^1 theta^j-1/2 cdot F_3,2[beginarrayrrr frac12& 1&1\ frac32& frac32 & endarray;theta]dthetaright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^l j binom-1/2j binom-1/2l-j cdot right.\
&&left.intlimits_0^pi/2 [sin(u)]^2j-1 2 cos(u)left( i textLi_2left(-e^i uright)-i textLi_2left(e^i uright)+u log left(frac1-e^i u1+e^i uright)right)duright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^lsumlimits_p=-j-1^j-1 j binom-1/2j binom-1/2l-j cdot frac(-1)^p+l2^2j-1[binom2j-1p+1 - binom2j-1p+j+1]cdotright.\
&&left.intlimits_1^imath z^p+1 left(log(z) log(frac1-z1+z)+Li_2(z)-Li_2(-z)right)dzright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C+right.\
&&left.
frac12
sumlimits_j=0^l-1sumlimits_p=-l-1,pneq -1^l-1
binom-3/2j binom-1/2l-1-j frac12^2j+1
[binom2j+1p+j+1-binom2j+1p+j+2]cdotright.\
&&left.
frac(-1)^p+lp+1
left((1+(-1)^p) cdot C - sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2right)right.\
&&left.
right)
endeqnarray
where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=exp(imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.
In summary we have shown that :
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
2 frac(1/2)^(lambda) (3/2)^(lambda)lambda! cdot lambda! cdot C + frac(3/2)^(lambda)lambda!cdot mathfrak A_lambda\
&&intlimits_0^pi/2 fraclog(1-cos(u))sqrt1-1/2 cos(u)^2=fracsqrtfrac2pi Gamma left(frac14right) (4 C+pi log (2))Gamma left(-frac14right)-\
&& sumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)\
&&intlimits_0^1 S(y^2) dy=-frac1pileft(right.\
&&left.
-frac2 Gamma left(frac54right) (pi (pi -2 log (4))-16 C)sqrtpi Gamma left(-frac14right)right.\
&&left.
-sqrt2cdotsumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)right.\
&&left.
right)
endeqnarray
where
begineqnarray
&&mathfrak A_lambda := \
&&left(
sumlimits_j=0^lambda-1 sumlimits_beginarrayr p=-lambda-1\pneq-1endarray^lambda-1
binom-3/2j binom-1/2lambda-1-j frac[binom2j+1p+j+1 - binom2j+1p+j+2]2^2 j+2
frac(-1)^p+lambda+1p+1cdot
sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2
right)
endeqnarray
Note : The remaining infinite sum converges quite quickly. Truncating the sum at the first one hundred terms produces a more than thirty digits accuracy. On the other hand the the series in question converges very slowly; one needs to take five thousand terms in order to get the first four decimal digits right.
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
$endgroup$
add a comment |
$begingroup$
This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach.
Denote :
beginequation
S(y):= sumlimits_n=0^infty binom4 n2 n binom2 nn fracy^2 n64^n
endequation
We use the good old identity:
beginequation
binom2 nn=(-4)^n cdot binom-frac12n
endequation
and we get
begineqnarray
S(y)&=& sumlimits_n=0^infty binom-frac12n cdot underbracebinom-frac122n_frac1pi intlimits_0^1 t^-1/2 (1-t)^2 n-1/2 dt cdot (-y^2)^n\
&=& frac1pi intlimits_0^1 frac1sqrtt(1-t)(1-y^2 t^2) dt
&=&
endeqnarray
Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives:
begineqnarray
intlimits_0^1 S(y^2) dy&=& frac1pi intlimits_0^1 fracF_2,1[frac14,frac12,frac54,t^2]sqrtt(1-t)dt\
&=&frac1pi intlimits_0^1 fracF[arcsin(sqrtt),-1]t sqrt1-t dt\
&=&-frac1piintlimits_0^1 frac2log(1-sqrt1-t)-log(t)2sqrtt(1-t^2)dt\
&=&-frac1pi left(2intlimits_0^1 fraclog(t)sqrtt(t-2)(t-1+sqrt2)(t-1-sqrt2)dt+pi^3/2 fracGamma(5/4)Gamma(3/4)right)\
&=&-frac1pi left(
2 sqrt2 intlimits_0^pi fraclog[sin(u/4)]sqrt3-cos(u)-
frac2 sqrtpi Gamma(5/4) (pi+4 log(2))Gamma(-1/4)
right)\
&=&-frac1pi left(
sqrt2 intlimits_0^pi fraclog[1-cos(u/2)]sqrt3-cos(u)du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)\
&=&
-frac1pi left(
sqrt2 intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)
endeqnarray
In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-sqrt1-t$.In the fifthe line I substituted for $t:=2 sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result:
begineqnarray
&&intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du =
-fracpi2sumlimits_lambda=0^infty
binom-1/2lambda (-frac12)^lambda cdot \
&&
left(frac1+lambda1+2lambda binom1/2+lambda-1/2(2 log(2)+H_lambda) + binomlambda-1/2 F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1] right)
endeqnarray
Now, the remaining hypergeometric functions $F_3,2$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity:
beginequation
frac(1+lambda)^(l)(3/2+lambda)^(l) = frac(3/2)^(lambda)1^(lambda) cdot frac1^(l)(3/2)^(l) cdot frac(l+1)^(lambda)(l+3/2)^(lambda)
endequation
and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity:
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C- sumlimits_j=1^lambda j binom-1/2+jlambda binomlambdaj (-1)^lambda-j cdot intlimits_0^1 theta^j-1/2 cdot F_3,2[beginarrayrrr frac12& 1&1\ frac32& frac32 & endarray;theta]dthetaright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^l j binom-1/2j binom-1/2l-j cdot right.\
&&left.intlimits_0^pi/2 [sin(u)]^2j-1 2 cos(u)left( i textLi_2left(-e^i uright)-i textLi_2left(e^i uright)+u log left(frac1-e^i u1+e^i uright)right)duright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^lsumlimits_p=-j-1^j-1 j binom-1/2j binom-1/2l-j cdot frac(-1)^p+l2^2j-1[binom2j-1p+1 - binom2j-1p+j+1]cdotright.\
&&left.intlimits_1^imath z^p+1 left(log(z) log(frac1-z1+z)+Li_2(z)-Li_2(-z)right)dzright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C+right.\
&&left.
frac12
sumlimits_j=0^l-1sumlimits_p=-l-1,pneq -1^l-1
binom-3/2j binom-1/2l-1-j frac12^2j+1
[binom2j+1p+j+1-binom2j+1p+j+2]cdotright.\
&&left.
frac(-1)^p+lp+1
left((1+(-1)^p) cdot C - sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2right)right.\
&&left.
right)
endeqnarray
where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=exp(imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.
In summary we have shown that :
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
2 frac(1/2)^(lambda) (3/2)^(lambda)lambda! cdot lambda! cdot C + frac(3/2)^(lambda)lambda!cdot mathfrak A_lambda\
&&intlimits_0^pi/2 fraclog(1-cos(u))sqrt1-1/2 cos(u)^2=fracsqrtfrac2pi Gamma left(frac14right) (4 C+pi log (2))Gamma left(-frac14right)-\
&& sumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)\
&&intlimits_0^1 S(y^2) dy=-frac1pileft(right.\
&&left.
-frac2 Gamma left(frac54right) (pi (pi -2 log (4))-16 C)sqrtpi Gamma left(-frac14right)right.\
&&left.
-sqrt2cdotsumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)right.\
&&left.
right)
endeqnarray
where
begineqnarray
&&mathfrak A_lambda := \
&&left(
sumlimits_j=0^lambda-1 sumlimits_beginarrayr p=-lambda-1\pneq-1endarray^lambda-1
binom-3/2j binom-1/2lambda-1-j frac[binom2j+1p+j+1 - binom2j+1p+j+2]2^2 j+2
frac(-1)^p+lambda+1p+1cdot
sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2
right)
endeqnarray
Note : The remaining infinite sum converges quite quickly. Truncating the sum at the first one hundred terms produces a more than thirty digits accuracy. On the other hand the the series in question converges very slowly; one needs to take five thousand terms in order to get the first four decimal digits right.
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
$endgroup$
add a comment |
$begingroup$
This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach.
Denote :
beginequation
S(y):= sumlimits_n=0^infty binom4 n2 n binom2 nn fracy^2 n64^n
endequation
We use the good old identity:
beginequation
binom2 nn=(-4)^n cdot binom-frac12n
endequation
and we get
begineqnarray
S(y)&=& sumlimits_n=0^infty binom-frac12n cdot underbracebinom-frac122n_frac1pi intlimits_0^1 t^-1/2 (1-t)^2 n-1/2 dt cdot (-y^2)^n\
&=& frac1pi intlimits_0^1 frac1sqrtt(1-t)(1-y^2 t^2) dt
&=&
endeqnarray
Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives:
begineqnarray
intlimits_0^1 S(y^2) dy&=& frac1pi intlimits_0^1 fracF_2,1[frac14,frac12,frac54,t^2]sqrtt(1-t)dt\
&=&frac1pi intlimits_0^1 fracF[arcsin(sqrtt),-1]t sqrt1-t dt\
&=&-frac1piintlimits_0^1 frac2log(1-sqrt1-t)-log(t)2sqrtt(1-t^2)dt\
&=&-frac1pi left(2intlimits_0^1 fraclog(t)sqrtt(t-2)(t-1+sqrt2)(t-1-sqrt2)dt+pi^3/2 fracGamma(5/4)Gamma(3/4)right)\
&=&-frac1pi left(
2 sqrt2 intlimits_0^pi fraclog[sin(u/4)]sqrt3-cos(u)-
frac2 sqrtpi Gamma(5/4) (pi+4 log(2))Gamma(-1/4)
right)\
&=&-frac1pi left(
sqrt2 intlimits_0^pi fraclog[1-cos(u/2)]sqrt3-cos(u)du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)\
&=&
-frac1pi left(
sqrt2 intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)
endeqnarray
In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-sqrt1-t$.In the fifthe line I substituted for $t:=2 sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result:
begineqnarray
&&intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du =
-fracpi2sumlimits_lambda=0^infty
binom-1/2lambda (-frac12)^lambda cdot \
&&
left(frac1+lambda1+2lambda binom1/2+lambda-1/2(2 log(2)+H_lambda) + binomlambda-1/2 F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1] right)
endeqnarray
Now, the remaining hypergeometric functions $F_3,2$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity:
beginequation
frac(1+lambda)^(l)(3/2+lambda)^(l) = frac(3/2)^(lambda)1^(lambda) cdot frac1^(l)(3/2)^(l) cdot frac(l+1)^(lambda)(l+3/2)^(lambda)
endequation
and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity:
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C- sumlimits_j=1^lambda j binom-1/2+jlambda binomlambdaj (-1)^lambda-j cdot intlimits_0^1 theta^j-1/2 cdot F_3,2[beginarrayrrr frac12& 1&1\ frac32& frac32 & endarray;theta]dthetaright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^l j binom-1/2j binom-1/2l-j cdot right.\
&&left.intlimits_0^pi/2 [sin(u)]^2j-1 2 cos(u)left( i textLi_2left(-e^i uright)-i textLi_2left(e^i uright)+u log left(frac1-e^i u1+e^i uright)right)duright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^lsumlimits_p=-j-1^j-1 j binom-1/2j binom-1/2l-j cdot frac(-1)^p+l2^2j-1[binom2j-1p+1 - binom2j-1p+j+1]cdotright.\
&&left.intlimits_1^imath z^p+1 left(log(z) log(frac1-z1+z)+Li_2(z)-Li_2(-z)right)dzright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C+right.\
&&left.
frac12
sumlimits_j=0^l-1sumlimits_p=-l-1,pneq -1^l-1
binom-3/2j binom-1/2l-1-j frac12^2j+1
[binom2j+1p+j+1-binom2j+1p+j+2]cdotright.\
&&left.
frac(-1)^p+lp+1
left((1+(-1)^p) cdot C - sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2right)right.\
&&left.
right)
endeqnarray
where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=exp(imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.
In summary we have shown that :
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
2 frac(1/2)^(lambda) (3/2)^(lambda)lambda! cdot lambda! cdot C + frac(3/2)^(lambda)lambda!cdot mathfrak A_lambda\
&&intlimits_0^pi/2 fraclog(1-cos(u))sqrt1-1/2 cos(u)^2=fracsqrtfrac2pi Gamma left(frac14right) (4 C+pi log (2))Gamma left(-frac14right)-\
&& sumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)\
&&intlimits_0^1 S(y^2) dy=-frac1pileft(right.\
&&left.
-frac2 Gamma left(frac54right) (pi (pi -2 log (4))-16 C)sqrtpi Gamma left(-frac14right)right.\
&&left.
-sqrt2cdotsumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)right.\
&&left.
right)
endeqnarray
where
begineqnarray
&&mathfrak A_lambda := \
&&left(
sumlimits_j=0^lambda-1 sumlimits_beginarrayr p=-lambda-1\pneq-1endarray^lambda-1
binom-3/2j binom-1/2lambda-1-j frac[binom2j+1p+j+1 - binom2j+1p+j+2]2^2 j+2
frac(-1)^p+lambda+1p+1cdot
sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2
right)
endeqnarray
Note : The remaining infinite sum converges quite quickly. Truncating the sum at the first one hundred terms produces a more than thirty digits accuracy. On the other hand the the series in question converges very slowly; one needs to take five thousand terms in order to get the first four decimal digits right.
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
$endgroup$
This is not going to be a full answer but I think I am able to contribute a little more compared to what has been presented above therefore I present my approach.
Denote :
beginequation
S(y):= sumlimits_n=0^infty binom4 n2 n binom2 nn fracy^2 n64^n
endequation
We use the good old identity:
beginequation
binom2 nn=(-4)^n cdot binom-frac12n
endequation
and we get
begineqnarray
S(y)&=& sumlimits_n=0^infty binom-frac12n cdot underbracebinom-frac122n_frac1pi intlimits_0^1 t^-1/2 (1-t)^2 n-1/2 dt cdot (-y^2)^n\
&=& frac1pi intlimits_0^1 frac1sqrtt(1-t)(1-y^2 t^2) dt
&=&
endeqnarray
Now in the last equation above we replace $y$ by $y^2$ and integrate over $y$ from zero to unity. This gives:
begineqnarray
intlimits_0^1 S(y^2) dy&=& frac1pi intlimits_0^1 fracF_2,1[frac14,frac12,frac54,t^2]sqrtt(1-t)dt\
&=&frac1pi intlimits_0^1 fracF[arcsin(sqrtt),-1]t sqrt1-t dt\
&=&-frac1piintlimits_0^1 frac2log(1-sqrt1-t)-log(t)2sqrtt(1-t^2)dt\
&=&-frac1pi left(2intlimits_0^1 fraclog(t)sqrtt(t-2)(t-1+sqrt2)(t-1-sqrt2)dt+pi^3/2 fracGamma(5/4)Gamma(3/4)right)\
&=&-frac1pi left(
2 sqrt2 intlimits_0^pi fraclog[sin(u/4)]sqrt3-cos(u)-
frac2 sqrtpi Gamma(5/4) (pi+4 log(2))Gamma(-1/4)
right)\
&=&-frac1pi left(
sqrt2 intlimits_0^pi fraclog[1-cos(u/2)]sqrt3-cos(u)du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)\
&=&
-frac1pi left(
sqrt2 intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du-
frac2 pi^3/2 Gamma(5/4)Gamma(-1/4)
right)
endeqnarray
In the first line above I expanded the square root in a series and integrated over $y$ term by term. In the second line above I went to Wolframs site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/09/19/02/ and looked up the closed form for the particular hypergeometric function. Here $F[phi,m]$ is the elliptic function of the first kind. In the third line I integrated by parts once and finally in the forth line I integrated the second term by using appropriate Euler sums and in the first term we substituted for $1-sqrt1-t$.In the fifthe line I substituted for $t:=2 sin(u)^2$ and in the sixth line I simplified the result using trigonometric half-angle identities.
The remaining integral can be actually simplified by expanding the integrand in a double series and then doing one of the sums. Here we just state the result:
begineqnarray
&&intlimits_0^pi/2 fraclog[1-cos(u)]sqrt1-1/2cos(u)^2du =
-fracpi2sumlimits_lambda=0^infty
binom-1/2lambda (-frac12)^lambda cdot \
&&
left(frac1+lambda1+2lambda binom1/2+lambda-1/2(2 log(2)+H_lambda) + binomlambda-1/2 F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1] right)
endeqnarray
Now, the remaining hypergeometric functions $F_3,2$ can also be simplified and expressed through Catalan numbers. We lack time to complete the whole calculations and therefore we have to leave this apart for the time being and complete this later.
Update: By using the following identity:
beginequation
frac(1+lambda)^(l)(3/2+lambda)^(l) = frac(3/2)^(lambda)1^(lambda) cdot frac1^(l)(3/2)^(l) cdot frac(l+1)^(lambda)(l+3/2)^(lambda)
endequation
and then by decomposing the last term on the righthand side into partial fractions in $l$ we easily get the following identity:
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C- sumlimits_j=1^lambda j binom-1/2+jlambda binomlambdaj (-1)^lambda-j cdot intlimits_0^1 theta^j-1/2 cdot F_3,2[beginarrayrrr frac12& 1&1\ frac32& frac32 & endarray;theta]dthetaright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^l j binom-1/2j binom-1/2l-j cdot right.\
&&left.intlimits_0^pi/2 [sin(u)]^2j-1 2 cos(u)left( i textLi_2left(-e^i uright)-i textLi_2left(e^i uright)+u log left(frac1-e^i u1+e^i uright)right)duright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C-(1-frac(-1)^l sqrtpiGamma[1/2-l]l!) fracimath pi^24-right.\
&&left.
(-1)^lsumlimits_j=1^lsumlimits_p=-j-1^j-1 j binom-1/2j binom-1/2l-j cdot frac(-1)^p+l2^2j-1[binom2j-1p+1 - binom2j-1p+j+1]cdotright.\
&&left.intlimits_1^imath z^p+1 left(log(z) log(frac1-z1+z)+Li_2(z)-Li_2(-z)right)dzright.\
&&left.
right)=\
&&frac(3/2)^(lambda)1^(lambda) cdot left(right.\
&&left.
2 C+right.\
&&left.
frac12
sumlimits_j=0^l-1sumlimits_p=-l-1,pneq -1^l-1
binom-3/2j binom-1/2l-1-j frac12^2j+1
[binom2j+1p+j+1-binom2j+1p+j+2]cdotright.\
&&left.
frac(-1)^p+lp+1
left((1+(-1)^p) cdot C - sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2right)right.\
&&left.
right)
endeqnarray
where in the second line we used the closed form expression from http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/02/01/07/0001/ and we changed the variables appropriately. In the third line we substituted for $z:=exp(imath u)$. In the fourth line we calculated the integrals using integration by parts and then simplified the result.
In summary we have shown that :
begineqnarray
&&F_3,2[beginarrayrrr frac12& 1&1+lambda\ frac32& frac32+lambda & endarray;1]=
2 frac(1/2)^(lambda) (3/2)^(lambda)lambda! cdot lambda! cdot C + frac(3/2)^(lambda)lambda!cdot mathfrak A_lambda\
&&intlimits_0^pi/2 fraclog(1-cos(u))sqrt1-1/2 cos(u)^2=fracsqrtfrac2pi Gamma left(frac14right) (4 C+pi log (2))Gamma left(-frac14right)-\
&& sumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)\
&&intlimits_0^1 S(y^2) dy=-frac1pileft(right.\
&&left.
-frac2 Gamma left(frac54right) (pi (pi -2 log (4))-16 C)sqrtpi Gamma left(-frac14right)right.\
&&left.
-sqrt2cdotsumlimits_lambda=0^infty binom-1/2lambda (frac12)^lambda left( fracpi4 binom-1/2lambda H_lambda+(-1)^lambda mathfrak A_lambdaright)right.\
&&left.
right)
endeqnarray
where
begineqnarray
&&mathfrak A_lambda := \
&&left(
sumlimits_j=0^lambda-1 sumlimits_beginarrayr p=-lambda-1\pneq-1endarray^lambda-1
binom-3/2j binom-1/2lambda-1-j frac[binom2j+1p+j+1 - binom2j+1p+j+2]2^2 j+2
frac(-1)^p+lambda+1p+1cdot
sumlimits_k=0^p cdot 1_pge 0-(p+2) 1_p<0 frac(-1)^k(2k+1)^2
right)
endeqnarray
Note : The remaining infinite sum converges quite quickly. Truncating the sum at the first one hundred terms produces a more than thirty digits accuracy. On the other hand the the series in question converges very slowly; one needs to take five thousand terms in order to get the first four decimal digits right.
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
edited Sep 28 '17 at 18:09
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
answered Sep 21 '17 at 12:33
PrzemoPrzemo
4,39311031
4,39311031
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$begingroup$
The hypergeometric term in question can also be expressed as a double integral of a relatively "nice" rational function: $_3F_2left(frac14,frac14,frac34;1,frac54;1right)=frac2sqrt2piint_0^1mathrmdxint_0^1mathrmdy,fracx^21-x^4+x^4y^4$.
$endgroup$
– David H
Sep 15 '17 at 10:27
$begingroup$
Which essentially boils down to $$ int_0^1fracarctan(z)+textarctanh(z)(1-z^4)^3/4,dz.$$ Maybe using some change of variables $varphi:(0,1)^2to (0,1)^2$ the integral $$ iint_(0,1)^2fracx^2,dx,dy1-x^4+x^4 y^4$$ becomes something with a nicer structure.
$endgroup$
– Jack D'Aurizio
Sep 15 '17 at 15:11
$begingroup$
I've mulled over the integral some more, and I think it may actually be a special case of an integral I asked about here. Some connections connections are made there with Appell functions that may prove useful somewhere, but for most part every way I try to attack this integral seems to be sending me in circles...
$endgroup$
– David H
Sep 25 '17 at 16:44