How to prove that $int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Mellin transform of $sin x$ aka $int^infty_0 x^s-1sin x dx $Solving $int_0^infty ln^m(x)sinleft(x^nright):dx$Solving used Real Based Methods: $int_0^x fract^kleft(t^n + aright)^m:dt$Proving $ int_0^infty fracln(t)sqrtte^-t mathrm dt=-sqrtpi(gamma+ln4)$How to prove $int_0^inftyfrace^-left(sqrtx-a/sqrtxright)^2sqrtxdx=sqrtpi,,a>0$?Solve: $int_0^infty left( cos left( … right)+sin left( … right) right) frace^-u^2(u^2+ fracalpha4t)^2 ~mathrmdu$Prove $Gamma^2(1/4)over 4sqrt2pi=prod_n=1^inftyleft(2n+1over 2nright)^(-1)^n+1$Proof $int_-infty^inftyfracdxleft(e^x+e^-x+e^ixsqrt3right)^2=frac13$How do I Evaluate the Integral $int_-infty ^infty frac1e^x^2+1 , dx$?Evaluate $int_0^infty sinleft(frac1xright)sinleft(x^2right),dx$What did I do wrong in calculating $int_-infty^inftyfracsinxxmathrmdx$ via $oint_gammafracsinzzmathrmdz$?How to prove $int_0^+infty fracsin(x)x^s = cos(fracpi s2) Gamma(1-s)$Using the result $g(omega)=frac1sqrt2pifrac2alphaalpha^2 + omega^2$ evaluate $int_-infty^inftyfrac11+x^2e^-ixdx$
Why is black pepper both grey and black?
What are the motives behind Cersei's orders given to Bronn?
Right-skewed distribution with mean equals to mode?
Java 8 stream max() function argument type Comparator vs Comparable
Why did the IBM 650 use bi-quinary?
I am not a queen, who am I?
How much radiation do nuclear physics experiments expose researchers to nowadays?
Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?
Letter Boxed validator
How to assign captions for two tables in LaTeX?
Does surprise arrest existing movement?
How can I fade player when goes inside or outside of the area?
How do I mention the quality of my school without bragging
When -s is used with third person singular. What's its use in this context?
Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?
Sorting numerically
The logistics of corpse disposal
Is there a way in Ruby to make just any one out of many keyword arguments required?
Why does Python start at index 1 when iterating an array backwards?
Output the ŋarâþ crîþ alphabet song without using (m)any letters
Stars Make Stars
How to motivate offshore teams and trust them to deliver?
What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
How to prove that $int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Mellin transform of $sin x$ aka $int^infty_0 x^s-1sin x dx $Solving $int_0^infty ln^m(x)sinleft(x^nright):dx$Solving used Real Based Methods: $int_0^x fract^kleft(t^n + aright)^m:dt$Proving $ int_0^infty fracln(t)sqrtte^-t mathrm dt=-sqrtpi(gamma+ln4)$How to prove $int_0^inftyfrace^-left(sqrtx-a/sqrtxright)^2sqrtxdx=sqrtpi,,a>0$?Solve: $int_0^infty left( cos left( … right)+sin left( … right) right) frace^-u^2(u^2+ fracalpha4t)^2 ~mathrmdu$Prove $Gamma^2(1/4)over 4sqrt2pi=prod_n=1^inftyleft(2n+1over 2nright)^(-1)^n+1$Proof $int_-infty^inftyfracdxleft(e^x+e^-x+e^ixsqrt3right)^2=frac13$How do I Evaluate the Integral $int_-infty ^infty frac1e^x^2+1 , dx$?Evaluate $int_0^infty sinleft(frac1xright)sinleft(x^2right),dx$What did I do wrong in calculating $int_-infty^inftyfracsinxxmathrmdx$ via $oint_gammafracsinzzmathrmdz$?How to prove $int_0^+infty fracsin(x)x^s = cos(fracpi s2) Gamma(1-s)$Using the result $g(omega)=frac1sqrt2pifrac2alphaalpha^2 + omega^2$ evaluate $int_-infty^inftyfrac11+x^2e^-ixdx$
$begingroup$
Wolfram Alpha provides
$$int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2tag1$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_0^inftyfrace^-t-e^-xttdt$$
I also know
$$int_0^inftysin(x^2)~dx=frac12int_0^inftyx^-1/2sin(x)~dx$$
Then,
$$beginalign
int_0^inftyln^2(x)sin(x^2)dx&=int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)sin(x^2)~dx\
&=frac12int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)fracsin(x)sqrtxdx\
&=frac12int_0^inftyint_0^inftyint_0^inftyfrace^-t-e^-xttfrace^-n-e^-xnnfracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-e^-xt)(e^-n-e^-xn)fracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-n-e^-xn-t-e^-xt-n+e^-xt-xn)fracsin(x)sqrtx~dx~dn~dt
endalign$$
What should I do next?
There is also a general case
$$int_0^inftyln^2(x^a)sin(x^2)dx=fraca^232sqrtfracpi2(2gamma-pi+ln16)^2tag2$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
$endgroup$
|
show 2 more comments
$begingroup$
Wolfram Alpha provides
$$int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2tag1$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_0^inftyfrace^-t-e^-xttdt$$
I also know
$$int_0^inftysin(x^2)~dx=frac12int_0^inftyx^-1/2sin(x)~dx$$
Then,
$$beginalign
int_0^inftyln^2(x)sin(x^2)dx&=int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)sin(x^2)~dx\
&=frac12int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)fracsin(x)sqrtxdx\
&=frac12int_0^inftyint_0^inftyint_0^inftyfrace^-t-e^-xttfrace^-n-e^-xnnfracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-e^-xt)(e^-n-e^-xn)fracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-n-e^-xn-t-e^-xt-n+e^-xt-xn)fracsin(x)sqrtx~dx~dn~dt
endalign$$
What should I do next?
There is also a general case
$$int_0^inftyln^2(x^a)sin(x^2)dx=fraca^232sqrtfracpi2(2gamma-pi+ln16)^2tag2$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
$endgroup$
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
2
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
$begingroup$
Wolfram Alpha provides
$$int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2tag1$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_0^inftyfrace^-t-e^-xttdt$$
I also know
$$int_0^inftysin(x^2)~dx=frac12int_0^inftyx^-1/2sin(x)~dx$$
Then,
$$beginalign
int_0^inftyln^2(x)sin(x^2)dx&=int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)sin(x^2)~dx\
&=frac12int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)fracsin(x)sqrtxdx\
&=frac12int_0^inftyint_0^inftyint_0^inftyfrace^-t-e^-xttfrace^-n-e^-xnnfracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-e^-xt)(e^-n-e^-xn)fracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-n-e^-xn-t-e^-xt-n+e^-xt-xn)fracsin(x)sqrtx~dx~dn~dt
endalign$$
What should I do next?
There is also a general case
$$int_0^inftyln^2(x^a)sin(x^2)dx=fraca^232sqrtfracpi2(2gamma-pi+ln16)^2tag2$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
$endgroup$
Wolfram Alpha provides
$$int_0^inftyln^2(x)sin(x^2)dx=frac132sqrtfracpi2(2gamma-pi+ln16)^2tag1$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_0^inftyfrace^-t-e^-xttdt$$
I also know
$$int_0^inftysin(x^2)~dx=frac12int_0^inftyx^-1/2sin(x)~dx$$
Then,
$$beginalign
int_0^inftyln^2(x)sin(x^2)dx&=int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)sin(x^2)~dx\
&=frac12int_0^inftyleft(int_0^inftyfrace^-t-e^-xttdtright)left(int_0^inftyfrace^-n-e^-xnndnright)fracsin(x)sqrtxdx\
&=frac12int_0^inftyint_0^inftyint_0^inftyfrace^-t-e^-xttfrace^-n-e^-xnnfracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-e^-xt)(e^-n-e^-xn)fracsin(x)sqrtx~dx~dn~dt\
&=frac12int_0^inftyfrac1tint_0^inftyfrac1nint_0^infty(e^-t-n-e^-xn-t-e^-xt-n+e^-xt-xn)fracsin(x)sqrtx~dx~dn~dt
endalign$$
What should I do next?
There is also a general case
$$int_0^inftyln^2(x^a)sin(x^2)dx=fraca^232sqrtfracpi2(2gamma-pi+ln16)^2tag2$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
calculus integration
edited Jan 26 at 20:18
user
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
edited Jan 26 at 20:18
user
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
edited Jan 26 at 20:18
user
6,57011031
edited Jan 26 at 20:18
user
6,57011031
edited Jan 26 at 20:18
user
6,57011031
6,57011031
asked Jan 26 at 19:48
LarryLarry
2,55031131
asked Jan 26 at 19:48
LarryLarry
2,55031131
asked Jan 26 at 19:48
LarryLarry
2,55031131
2,55031131
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
2
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
2
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
2
2
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
6 Answers
6
active
oldest
votes
$begingroup$
$$I=int_0^inftyln^2(x)sin(x^2)dx oversetx^2=t=int_0^infty frac12sqrt t ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^-1/2sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^s-1sin x dx= Gamma(s) sinleft(fracpi s2right)$$
$$Rightarrow I=frac18fracd^2ds^2Gamma(s) sinleft(fracpi s2right)bigg|_s=frac12$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow fracddsGamma(s) sinleft(fracpi s2right)=Gamma(s)psi(s)sinleft(fracpi s2right) +fracpi2Gamma(s)cosleft(fracpi s2right)$$
$$Rightarrow fracd^2ds^2Gamma(s) sinleft(fracpi s2right)=fracddsGamma(s)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)+Gamma(s)left(psi_1(x)sinleft(fracpi s2right)+fracpi2Gamma(s)cosleft(fracpi s2right)-fracpi^24sinleft(fracpi s2right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrtpi$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=fracpi^22$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,87511062
$endgroup$
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
add a comment |
$begingroup$
We have
$$ F(alpha)=int_0^+infty x^alpha sin(x^2),dx = frac12int_0^+infty x^alpha/2-1sin(x),dx\=frac12Gamma(1-alpha/2)int_0^+infty fracdss^alpha/2(s^2+1) $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac12,Gammaleft(frac1+alpha2right)sinleft(fracpi4(1+alpha)right) $$
for any $alpha$ such that $textRe(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_alphato 0fracd^2dalpha^2$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac12$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac18int_0^infty fracln^2(x)sin(x)sqrtxdx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^p-1sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
$endgroup$
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^inftylog^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^alog x)=alog x$,
$$F(a)=a^2int_0^inftylog^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrmD^2_s=frac12Gamma(s)sinfracpi s2=frac132sqrtfracpi2(2gamma-pi+log16)^2$$
So
$$F(a)=fraca^232sqrtfracpi2(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
$endgroup$
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx
endequation
We first observe that:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2int_0^infty x^ksinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2 H(k)
endequation
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
beginequation
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
endequation
(This is allowable through the Dominated Convergence Theorem). Thus:
beginequation
H(k) = lim_trightarrow 1^+ J(t; k)
endequation
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
beginalign
mathscrL_tleft[J(t;k) right] &= int_0^infty x^kmathscrL_tleft[sinleft(tx^2right)right]:dx = int_0^infty fracx^k + 2s^2 + x^4:dx
endalign
As I address here we find this becomes:
beginalign
mathscrL_tleft[J(t;k) right] &= frac14cdot left(s^2right)^frack + 2 + 12 - 1 cdot Bleft(1 - frack + 2 + 1 4, frack + 2 + 1 4 right) = frac14 s^frack - 12 Bleft(1 - frack + 34 , frack + 34right)
endalign
Using the relationship between the Gamma and Beta Function we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 Gammaleft(1 - frack + 34right) Gammaleft( frack + 34right)
endequation
Using Euler's Reflection Formula we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)
endequation
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
beginequation
I = lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right]
endequation
In this process we solve for $H(k)$ using
beginequation
H(k) = lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]
endequation
Thus, our definition of $I$ becomes:
beginalign
I &= lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right] = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]right] \
&= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2mathscrL_tleft[J(t; k)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right]
endalign
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
beginalign
I &= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ fracpi4left( frac3pi^28sqrt2sqrts + fracln^2(s)2sqrt2sqrts + fracpiln(s)2sqrt2sqrtsright)right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 mathscrL_s^-1left[ frac1sqrtsright] + fracpi8sqrt2 mathscrL_s^-1left[ fracln^2(s)sqrtsright]+ fracpi^28sqrt2 mathscrL_s^-1left[ fracln(s)sqrtsright]right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 left[ frac1sqrtpisqrttright] + fracpi32sqrt2 left[ frac
left(psi^(0)left(frac12right)-ln(t)right)^2 -fracpi^22sqrtpisqrttright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)-ln(t)sqrtpisqrttright]right] \
&= frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
psi^(0)left(frac12right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)sqrtpiright]
endalign
Noting that
beginequation
psi^(0)left(frac12right) = -gamma - 2ln(2)
endequation
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
beginalign
I = frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
left(gamma + 2ln(2)right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
gamma - 2ln(2)sqrtpiright] = frac132sqrtfracpi2(2gamma-pi+4ln2)^2
endalign
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsR > 0$ and $dsnu in pars0,1$:
beginalign
&bbox[10px,#ffd]int_0^Rx^nuexpparsic x^2dd x
\[5mm] = &
-int_0^pi/4parsRexpoic theta^nu
expparsic R^2expo2icthetaRexpoicthetaic,ddtheta -
int_R^0parsrexpoicpi/4^nu
expparsicbracksrexpoicpi/4^2expoicpi/4,dd r
\[8mm] = &
-overbraceR^nu + 1,icint_0^pi/4
expparsicbracksnutheta + R^2cospars2theta + theta
exppars-R^2sinpars2thetaddtheta^dsequiv mcIparsR,nu
\[2mm] + &
expoicparsnu + 1pi/4int_0^Rr^nuexpo-r^2dd r
endalign
Since $dsnu in pars0,1$, note that
beginalign
0 & < vertsmcIparsR,nu <
R^nu + 1int_0^pi/4expo-4R^2theta/piddtheta =
pi over 4,1 - exppars-R^2 over R^1 - nu
,,,stackrelmrmas R to inftyLARGEto,,,
colorredlarge 0
endalign
Then,
beginalign
&bbox[10px,#ffd]int_0^inftyx^nusinparsx^2dd x =
sinparsbracksnu + 1,pi over 4
int_0^inftyr^nuexpo-r^2dd r
\[5mm] stackrelr^2 mapsto r=,,,&
1 over 2,sinparsbracksnu + 1,pi over 4
int_0^inftyr^nu/2 - 1/2expo-rdd r =
1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
endalign
and
beginalign
&bbox[10px,#ffd]int_0^inftyln^2parsxsinparsx^2dd x =
lim_nu to 0^+
totald[2]nu
braces1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
\[5mm] = &
bbx1 over 32rootpi over 2
bracksvphantomLarge A2gamma - pi + lnpars16^2 approx 0.0242
endalign
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088677%2fhow-to-prove-that-int-0-infty-ln2x-sinx2dx-frac132-sqrt-frac%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$I=int_0^inftyln^2(x)sin(x^2)dx oversetx^2=t=int_0^infty frac12sqrt t ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^-1/2sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^s-1sin x dx= Gamma(s) sinleft(fracpi s2right)$$
$$Rightarrow I=frac18fracd^2ds^2Gamma(s) sinleft(fracpi s2right)bigg|_s=frac12$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow fracddsGamma(s) sinleft(fracpi s2right)=Gamma(s)psi(s)sinleft(fracpi s2right) +fracpi2Gamma(s)cosleft(fracpi s2right)$$
$$Rightarrow fracd^2ds^2Gamma(s) sinleft(fracpi s2right)=fracddsGamma(s)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)+Gamma(s)left(psi_1(x)sinleft(fracpi s2right)+fracpi2Gamma(s)cosleft(fracpi s2right)-fracpi^24sinleft(fracpi s2right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrtpi$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=fracpi^22$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,87511062
$endgroup$
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
add a comment |
$begingroup$
$$I=int_0^inftyln^2(x)sin(x^2)dx oversetx^2=t=int_0^infty frac12sqrt t ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^-1/2sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^s-1sin x dx= Gamma(s) sinleft(fracpi s2right)$$
$$Rightarrow I=frac18fracd^2ds^2Gamma(s) sinleft(fracpi s2right)bigg|_s=frac12$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow fracddsGamma(s) sinleft(fracpi s2right)=Gamma(s)psi(s)sinleft(fracpi s2right) +fracpi2Gamma(s)cosleft(fracpi s2right)$$
$$Rightarrow fracd^2ds^2Gamma(s) sinleft(fracpi s2right)=fracddsGamma(s)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)+Gamma(s)left(psi_1(x)sinleft(fracpi s2right)+fracpi2Gamma(s)cosleft(fracpi s2right)-fracpi^24sinleft(fracpi s2right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrtpi$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=fracpi^22$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,87511062
$endgroup$
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
add a comment |
$begingroup$
$$I=int_0^inftyln^2(x)sin(x^2)dx oversetx^2=t=int_0^infty frac12sqrt t ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^-1/2sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^s-1sin x dx= Gamma(s) sinleft(fracpi s2right)$$
$$Rightarrow I=frac18fracd^2ds^2Gamma(s) sinleft(fracpi s2right)bigg|_s=frac12$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow fracddsGamma(s) sinleft(fracpi s2right)=Gamma(s)psi(s)sinleft(fracpi s2right) +fracpi2Gamma(s)cosleft(fracpi s2right)$$
$$Rightarrow fracd^2ds^2Gamma(s) sinleft(fracpi s2right)=fracddsGamma(s)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)+Gamma(s)left(psi_1(x)sinleft(fracpi s2right)+fracpi2Gamma(s)cosleft(fracpi s2right)-fracpi^24sinleft(fracpi s2right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrtpi$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=fracpi^22$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,87511062
$endgroup$
$$I=int_0^inftyln^2(x)sin(x^2)dx oversetx^2=t=int_0^infty frac12sqrt t ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^-1/2sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^s-1sin x dx= Gamma(s) sinleft(fracpi s2right)$$
$$Rightarrow I=frac18fracd^2ds^2Gamma(s) sinleft(fracpi s2right)bigg|_s=frac12$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow fracddsGamma(s) sinleft(fracpi s2right)=Gamma(s)psi(s)sinleft(fracpi s2right) +fracpi2Gamma(s)cosleft(fracpi s2right)$$
$$Rightarrow fracd^2ds^2Gamma(s) sinleft(fracpi s2right)=fracddsGamma(s)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(fracpi s2right)+fracpi2cosleft(fracpi s2right)right)+Gamma(s)left(psi_1(x)sinleft(fracpi s2right)+fracpi2Gamma(s)cosleft(fracpi s2right)-fracpi^24sinleft(fracpi s2right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrtpi$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=fracpi^22$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,87511062
answered Jan 26 at 20:24
ZackyZacky
7,87511062
answered Jan 26 at 20:24
ZackyZacky
7,87511062
answered Jan 26 at 20:24
ZackyZacky
7,87511062
7,87511062
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
add a comment |
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
2
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– user150203
Jan 28 at 9:09
add a comment |
$begingroup$
We have
$$ F(alpha)=int_0^+infty x^alpha sin(x^2),dx = frac12int_0^+infty x^alpha/2-1sin(x),dx\=frac12Gamma(1-alpha/2)int_0^+infty fracdss^alpha/2(s^2+1) $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac12,Gammaleft(frac1+alpha2right)sinleft(fracpi4(1+alpha)right) $$
for any $alpha$ such that $textRe(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_alphato 0fracd^2dalpha^2$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac12$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
We have
$$ F(alpha)=int_0^+infty x^alpha sin(x^2),dx = frac12int_0^+infty x^alpha/2-1sin(x),dx\=frac12Gamma(1-alpha/2)int_0^+infty fracdss^alpha/2(s^2+1) $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac12,Gammaleft(frac1+alpha2right)sinleft(fracpi4(1+alpha)right) $$
for any $alpha$ such that $textRe(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_alphato 0fracd^2dalpha^2$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac12$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
We have
$$ F(alpha)=int_0^+infty x^alpha sin(x^2),dx = frac12int_0^+infty x^alpha/2-1sin(x),dx\=frac12Gamma(1-alpha/2)int_0^+infty fracdss^alpha/2(s^2+1) $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac12,Gammaleft(frac1+alpha2right)sinleft(fracpi4(1+alpha)right) $$
for any $alpha$ such that $textRe(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_alphato 0fracd^2dalpha^2$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac12$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
We have
$$ F(alpha)=int_0^+infty x^alpha sin(x^2),dx = frac12int_0^+infty x^alpha/2-1sin(x),dx\=frac12Gamma(1-alpha/2)int_0^+infty fracdss^alpha/2(s^2+1) $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac12,Gammaleft(frac1+alpha2right)sinleft(fracpi4(1+alpha)right) $$
for any $alpha$ such that $textRe(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_alphato 0fracd^2dalpha^2$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac12$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
292k33284673
292k33284673
add a comment |
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac18int_0^infty fracln^2(x)sin(x)sqrtxdx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^p-1sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
$endgroup$
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac18int_0^infty fracln^2(x)sin(x)sqrtxdx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^p-1sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
$endgroup$
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac18int_0^infty fracln^2(x)sin(x)sqrtxdx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^p-1sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
$endgroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac18int_0^infty fracln^2(x)sin(x)sqrtxdx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^p-1sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841112
23k841112
add a comment |
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^inftylog^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^alog x)=alog x$,
$$F(a)=a^2int_0^inftylog^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrmD^2_s=frac12Gamma(s)sinfracpi s2=frac132sqrtfracpi2(2gamma-pi+log16)^2$$
So
$$F(a)=fraca^232sqrtfracpi2(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
$endgroup$
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^inftylog^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^alog x)=alog x$,
$$F(a)=a^2int_0^inftylog^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrmD^2_s=frac12Gamma(s)sinfracpi s2=frac132sqrtfracpi2(2gamma-pi+log16)^2$$
So
$$F(a)=fraca^232sqrtfracpi2(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
$endgroup$
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^inftylog^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^alog x)=alog x$,
$$F(a)=a^2int_0^inftylog^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrmD^2_s=frac12Gamma(s)sinfracpi s2=frac132sqrtfracpi2(2gamma-pi+log16)^2$$
So
$$F(a)=fraca^232sqrtfracpi2(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
$endgroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^inftylog^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^alog x)=alog x$,
$$F(a)=a^2int_0^inftylog^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrmD^2_s=frac12Gamma(s)sinfracpi s2=frac132sqrtfracpi2(2gamma-pi+log16)^2$$
So
$$F(a)=fraca^232sqrtfracpi2(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
answered Jan 26 at 21:05
clathratusclathratus
5,1441439
5,1441439
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx
endequation
We first observe that:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2int_0^infty x^ksinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2 H(k)
endequation
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
beginequation
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
endequation
(This is allowable through the Dominated Convergence Theorem). Thus:
beginequation
H(k) = lim_trightarrow 1^+ J(t; k)
endequation
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
beginalign
mathscrL_tleft[J(t;k) right] &= int_0^infty x^kmathscrL_tleft[sinleft(tx^2right)right]:dx = int_0^infty fracx^k + 2s^2 + x^4:dx
endalign
As I address here we find this becomes:
beginalign
mathscrL_tleft[J(t;k) right] &= frac14cdot left(s^2right)^frack + 2 + 12 - 1 cdot Bleft(1 - frack + 2 + 1 4, frack + 2 + 1 4 right) = frac14 s^frack - 12 Bleft(1 - frack + 34 , frack + 34right)
endalign
Using the relationship between the Gamma and Beta Function we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 Gammaleft(1 - frack + 34right) Gammaleft( frack + 34right)
endequation
Using Euler's Reflection Formula we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)
endequation
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
beginequation
I = lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right]
endequation
In this process we solve for $H(k)$ using
beginequation
H(k) = lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]
endequation
Thus, our definition of $I$ becomes:
beginalign
I &= lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right] = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]right] \
&= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2mathscrL_tleft[J(t; k)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right]
endalign
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
beginalign
I &= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ fracpi4left( frac3pi^28sqrt2sqrts + fracln^2(s)2sqrt2sqrts + fracpiln(s)2sqrt2sqrtsright)right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 mathscrL_s^-1left[ frac1sqrtsright] + fracpi8sqrt2 mathscrL_s^-1left[ fracln^2(s)sqrtsright]+ fracpi^28sqrt2 mathscrL_s^-1left[ fracln(s)sqrtsright]right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 left[ frac1sqrtpisqrttright] + fracpi32sqrt2 left[ frac
left(psi^(0)left(frac12right)-ln(t)right)^2 -fracpi^22sqrtpisqrttright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)-ln(t)sqrtpisqrttright]right] \
&= frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
psi^(0)left(frac12right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)sqrtpiright]
endalign
Noting that
beginequation
psi^(0)left(frac12right) = -gamma - 2ln(2)
endequation
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
beginalign
I = frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
left(gamma + 2ln(2)right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
gamma - 2ln(2)sqrtpiright] = frac132sqrtfracpi2(2gamma-pi+4ln2)^2
endalign
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx
endequation
We first observe that:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2int_0^infty x^ksinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2 H(k)
endequation
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
beginequation
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
endequation
(This is allowable through the Dominated Convergence Theorem). Thus:
beginequation
H(k) = lim_trightarrow 1^+ J(t; k)
endequation
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
beginalign
mathscrL_tleft[J(t;k) right] &= int_0^infty x^kmathscrL_tleft[sinleft(tx^2right)right]:dx = int_0^infty fracx^k + 2s^2 + x^4:dx
endalign
As I address here we find this becomes:
beginalign
mathscrL_tleft[J(t;k) right] &= frac14cdot left(s^2right)^frack + 2 + 12 - 1 cdot Bleft(1 - frack + 2 + 1 4, frack + 2 + 1 4 right) = frac14 s^frack - 12 Bleft(1 - frack + 34 , frack + 34right)
endalign
Using the relationship between the Gamma and Beta Function we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 Gammaleft(1 - frack + 34right) Gammaleft( frack + 34right)
endequation
Using Euler's Reflection Formula we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)
endequation
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
beginequation
I = lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right]
endequation
In this process we solve for $H(k)$ using
beginequation
H(k) = lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]
endequation
Thus, our definition of $I$ becomes:
beginalign
I &= lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right] = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]right] \
&= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2mathscrL_tleft[J(t; k)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right]
endalign
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
beginalign
I &= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ fracpi4left( frac3pi^28sqrt2sqrts + fracln^2(s)2sqrt2sqrts + fracpiln(s)2sqrt2sqrtsright)right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 mathscrL_s^-1left[ frac1sqrtsright] + fracpi8sqrt2 mathscrL_s^-1left[ fracln^2(s)sqrtsright]+ fracpi^28sqrt2 mathscrL_s^-1left[ fracln(s)sqrtsright]right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 left[ frac1sqrtpisqrttright] + fracpi32sqrt2 left[ frac
left(psi^(0)left(frac12right)-ln(t)right)^2 -fracpi^22sqrtpisqrttright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)-ln(t)sqrtpisqrttright]right] \
&= frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
psi^(0)left(frac12right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)sqrtpiright]
endalign
Noting that
beginequation
psi^(0)left(frac12right) = -gamma - 2ln(2)
endequation
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
beginalign
I = frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
left(gamma + 2ln(2)right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
gamma - 2ln(2)sqrtpiright] = frac132sqrtfracpi2(2gamma-pi+4ln2)^2
endalign
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx
endequation
We first observe that:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2int_0^infty x^ksinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2 H(k)
endequation
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
beginequation
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
endequation
(This is allowable through the Dominated Convergence Theorem). Thus:
beginequation
H(k) = lim_trightarrow 1^+ J(t; k)
endequation
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
beginalign
mathscrL_tleft[J(t;k) right] &= int_0^infty x^kmathscrL_tleft[sinleft(tx^2right)right]:dx = int_0^infty fracx^k + 2s^2 + x^4:dx
endalign
As I address here we find this becomes:
beginalign
mathscrL_tleft[J(t;k) right] &= frac14cdot left(s^2right)^frack + 2 + 12 - 1 cdot Bleft(1 - frack + 2 + 1 4, frack + 2 + 1 4 right) = frac14 s^frack - 12 Bleft(1 - frack + 34 , frack + 34right)
endalign
Using the relationship between the Gamma and Beta Function we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 Gammaleft(1 - frack + 34right) Gammaleft( frack + 34right)
endequation
Using Euler's Reflection Formula we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)
endequation
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
beginequation
I = lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right]
endequation
In this process we solve for $H(k)$ using
beginequation
H(k) = lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]
endequation
Thus, our definition of $I$ becomes:
beginalign
I &= lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right] = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]right] \
&= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2mathscrL_tleft[J(t; k)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right]
endalign
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
beginalign
I &= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ fracpi4left( frac3pi^28sqrt2sqrts + fracln^2(s)2sqrt2sqrts + fracpiln(s)2sqrt2sqrtsright)right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 mathscrL_s^-1left[ frac1sqrtsright] + fracpi8sqrt2 mathscrL_s^-1left[ fracln^2(s)sqrtsright]+ fracpi^28sqrt2 mathscrL_s^-1left[ fracln(s)sqrtsright]right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 left[ frac1sqrtpisqrttright] + fracpi32sqrt2 left[ frac
left(psi^(0)left(frac12right)-ln(t)right)^2 -fracpi^22sqrtpisqrttright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)-ln(t)sqrtpisqrttright]right] \
&= frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
psi^(0)left(frac12right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)sqrtpiright]
endalign
Noting that
beginequation
psi^(0)left(frac12right) = -gamma - 2ln(2)
endequation
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
beginalign
I = frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
left(gamma + 2ln(2)right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
gamma - 2ln(2)sqrtpiright] = frac132sqrtfracpi2(2gamma-pi+4ln2)^2
endalign
$endgroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx
endequation
We first observe that:
beginequation
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2int_0^infty x^ksinleft(x^2right):dx = lim_krightarrow 0^+ fracd^2dk^2 H(k)
endequation
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
beginequation
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
endequation
(This is allowable through the Dominated Convergence Theorem). Thus:
beginequation
H(k) = lim_trightarrow 1^+ J(t; k)
endequation
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
beginalign
mathscrL_tleft[J(t;k) right] &= int_0^infty x^kmathscrL_tleft[sinleft(tx^2right)right]:dx = int_0^infty fracx^k + 2s^2 + x^4:dx
endalign
As I address here we find this becomes:
beginalign
mathscrL_tleft[J(t;k) right] &= frac14cdot left(s^2right)^frack + 2 + 12 - 1 cdot Bleft(1 - frack + 2 + 1 4, frack + 2 + 1 4 right) = frac14 s^frack - 12 Bleft(1 - frack + 34 , frack + 34right)
endalign
Using the relationship between the Gamma and Beta Function we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 Gammaleft(1 - frack + 34right) Gammaleft( frack + 34right)
endequation
Using Euler's Reflection Formula we find:
beginequation
mathscrL_tleft[J(t;k) right] = frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)
endequation
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
beginequation
I = lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right]
endequation
In this process we solve for $H(k)$ using
beginequation
H(k) = lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]
endequation
Thus, our definition of $I$ becomes:
beginalign
I &= lim_krightarrow 0^+ fracd^2dk^2 H(k) = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ J(t;k)right] = lim_krightarrow 0^+ fracd^2dk^2left[ lim_trightarrow 1^+ mathscrL_s^-1left[mathscrL_tleft[J(t; k)right]right]right] \
&= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2mathscrL_tleft[J(t; k)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right]
endalign
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
beginalign
I &= lim_trightarrow 1^+ mathscrL_s^-1left[ lim_krightarrow 0^+ fracd^2dk^2left[ frac14 s^frack - 12 fracpisinleft(pileft(frack + 34right) right)right]right] = lim_trightarrow 1^+ mathscrL_s^-1left[ fracpi4left( frac3pi^28sqrt2sqrts + fracln^2(s)2sqrt2sqrts + fracpiln(s)2sqrt2sqrtsright)right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 mathscrL_s^-1left[ frac1sqrtsright] + fracpi8sqrt2 mathscrL_s^-1left[ fracln^2(s)sqrtsright]+ fracpi^28sqrt2 mathscrL_s^-1left[ fracln(s)sqrtsright]right] \
&= lim_trightarrow 1^+ left[ frac3pi^332sqrt2 left[ frac1sqrtpisqrttright] + fracpi32sqrt2 left[ frac
left(psi^(0)left(frac12right)-ln(t)right)^2 -fracpi^22sqrtpisqrttright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)-ln(t)sqrtpisqrttright]right] \
&= frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
psi^(0)left(frac12right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
psi^(0)left(frac12right)sqrtpiright]
endalign
Noting that
beginequation
psi^(0)left(frac12right) = -gamma - 2ln(2)
endequation
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
beginalign
I = frac3pi^332sqrt2 left[ frac1sqrtpiright] + fracpi32sqrt2 left[ frac
left(gamma + 2ln(2)right)^2 -fracpi^22sqrtpiright]+ fracpi^216sqrt2 left[ frac
gamma - 2ln(2)sqrtpiright] = frac132sqrtfracpi2(2gamma-pi+4ln2)^2
endalign
edited Jan 31 at 12:44
answered Jan 28 at 12:51
user150203
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
add a comment |
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
$begingroup$
@clathratus You definitely can
$endgroup$
– user150203
Feb 7 at 0:09
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsR > 0$ and $dsnu in pars0,1$:
beginalign
&bbox[10px,#ffd]int_0^Rx^nuexpparsic x^2dd x
\[5mm] = &
-int_0^pi/4parsRexpoic theta^nu
expparsic R^2expo2icthetaRexpoicthetaic,ddtheta -
int_R^0parsrexpoicpi/4^nu
expparsicbracksrexpoicpi/4^2expoicpi/4,dd r
\[8mm] = &
-overbraceR^nu + 1,icint_0^pi/4
expparsicbracksnutheta + R^2cospars2theta + theta
exppars-R^2sinpars2thetaddtheta^dsequiv mcIparsR,nu
\[2mm] + &
expoicparsnu + 1pi/4int_0^Rr^nuexpo-r^2dd r
endalign
Since $dsnu in pars0,1$, note that
beginalign
0 & < vertsmcIparsR,nu <
R^nu + 1int_0^pi/4expo-4R^2theta/piddtheta =
pi over 4,1 - exppars-R^2 over R^1 - nu
,,,stackrelmrmas R to inftyLARGEto,,,
colorredlarge 0
endalign
Then,
beginalign
&bbox[10px,#ffd]int_0^inftyx^nusinparsx^2dd x =
sinparsbracksnu + 1,pi over 4
int_0^inftyr^nuexpo-r^2dd r
\[5mm] stackrelr^2 mapsto r=,,,&
1 over 2,sinparsbracksnu + 1,pi over 4
int_0^inftyr^nu/2 - 1/2expo-rdd r =
1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
endalign
and
beginalign
&bbox[10px,#ffd]int_0^inftyln^2parsxsinparsx^2dd x =
lim_nu to 0^+
totald[2]nu
braces1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
\[5mm] = &
bbx1 over 32rootpi over 2
bracksvphantomLarge A2gamma - pi + lnpars16^2 approx 0.0242
endalign
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
$endgroup$
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsR > 0$ and $dsnu in pars0,1$:
beginalign
&bbox[10px,#ffd]int_0^Rx^nuexpparsic x^2dd x
\[5mm] = &
-int_0^pi/4parsRexpoic theta^nu
expparsic R^2expo2icthetaRexpoicthetaic,ddtheta -
int_R^0parsrexpoicpi/4^nu
expparsicbracksrexpoicpi/4^2expoicpi/4,dd r
\[8mm] = &
-overbraceR^nu + 1,icint_0^pi/4
expparsicbracksnutheta + R^2cospars2theta + theta
exppars-R^2sinpars2thetaddtheta^dsequiv mcIparsR,nu
\[2mm] + &
expoicparsnu + 1pi/4int_0^Rr^nuexpo-r^2dd r
endalign
Since $dsnu in pars0,1$, note that
beginalign
0 & < vertsmcIparsR,nu <
R^nu + 1int_0^pi/4expo-4R^2theta/piddtheta =
pi over 4,1 - exppars-R^2 over R^1 - nu
,,,stackrelmrmas R to inftyLARGEto,,,
colorredlarge 0
endalign
Then,
beginalign
&bbox[10px,#ffd]int_0^inftyx^nusinparsx^2dd x =
sinparsbracksnu + 1,pi over 4
int_0^inftyr^nuexpo-r^2dd r
\[5mm] stackrelr^2 mapsto r=,,,&
1 over 2,sinparsbracksnu + 1,pi over 4
int_0^inftyr^nu/2 - 1/2expo-rdd r =
1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
endalign
and
beginalign
&bbox[10px,#ffd]int_0^inftyln^2parsxsinparsx^2dd x =
lim_nu to 0^+
totald[2]nu
braces1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
\[5mm] = &
bbx1 over 32rootpi over 2
bracksvphantomLarge A2gamma - pi + lnpars16^2 approx 0.0242
endalign
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
$endgroup$
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsR > 0$ and $dsnu in pars0,1$:
beginalign
&bbox[10px,#ffd]int_0^Rx^nuexpparsic x^2dd x
\[5mm] = &
-int_0^pi/4parsRexpoic theta^nu
expparsic R^2expo2icthetaRexpoicthetaic,ddtheta -
int_R^0parsrexpoicpi/4^nu
expparsicbracksrexpoicpi/4^2expoicpi/4,dd r
\[8mm] = &
-overbraceR^nu + 1,icint_0^pi/4
expparsicbracksnutheta + R^2cospars2theta + theta
exppars-R^2sinpars2thetaddtheta^dsequiv mcIparsR,nu
\[2mm] + &
expoicparsnu + 1pi/4int_0^Rr^nuexpo-r^2dd r
endalign
Since $dsnu in pars0,1$, note that
beginalign
0 & < vertsmcIparsR,nu <
R^nu + 1int_0^pi/4expo-4R^2theta/piddtheta =
pi over 4,1 - exppars-R^2 over R^1 - nu
,,,stackrelmrmas R to inftyLARGEto,,,
colorredlarge 0
endalign
Then,
beginalign
&bbox[10px,#ffd]int_0^inftyx^nusinparsx^2dd x =
sinparsbracksnu + 1,pi over 4
int_0^inftyr^nuexpo-r^2dd r
\[5mm] stackrelr^2 mapsto r=,,,&
1 over 2,sinparsbracksnu + 1,pi over 4
int_0^inftyr^nu/2 - 1/2expo-rdd r =
1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
endalign
and
beginalign
&bbox[10px,#ffd]int_0^inftyln^2parsxsinparsx^2dd x =
lim_nu to 0^+
totald[2]nu
braces1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
\[5mm] = &
bbx1 over 32rootpi over 2
bracksvphantomLarge A2gamma - pi + lnpars16^2 approx 0.0242
endalign
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
$endgroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsR > 0$ and $dsnu in pars0,1$:
beginalign
&bbox[10px,#ffd]int_0^Rx^nuexpparsic x^2dd x
\[5mm] = &
-int_0^pi/4parsRexpoic theta^nu
expparsic R^2expo2icthetaRexpoicthetaic,ddtheta -
int_R^0parsrexpoicpi/4^nu
expparsicbracksrexpoicpi/4^2expoicpi/4,dd r
\[8mm] = &
-overbraceR^nu + 1,icint_0^pi/4
expparsicbracksnutheta + R^2cospars2theta + theta
exppars-R^2sinpars2thetaddtheta^dsequiv mcIparsR,nu
\[2mm] + &
expoicparsnu + 1pi/4int_0^Rr^nuexpo-r^2dd r
endalign
Since $dsnu in pars0,1$, note that
beginalign
0 & < vertsmcIparsR,nu <
R^nu + 1int_0^pi/4expo-4R^2theta/piddtheta =
pi over 4,1 - exppars-R^2 over R^1 - nu
,,,stackrelmrmas R to inftyLARGEto,,,
colorredlarge 0
endalign
Then,
beginalign
&bbox[10px,#ffd]int_0^inftyx^nusinparsx^2dd x =
sinparsbracksnu + 1,pi over 4
int_0^inftyr^nuexpo-r^2dd r
\[5mm] stackrelr^2 mapsto r=,,,&
1 over 2,sinparsbracksnu + 1,pi over 4
int_0^inftyr^nu/2 - 1/2expo-rdd r =
1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
endalign
and
beginalign
&bbox[10px,#ffd]int_0^inftyln^2parsxsinparsx^2dd x =
lim_nu to 0^+
totald[2]nu
braces1 over 2,sinparsbracksnu + 1,pi over 4
Gammaparsnu over 2 + 1 over 2
\[5mm] = &
bbx1 over 32rootpi over 2
bracksvphantomLarge A2gamma - pi + lnpars16^2 approx 0.0242
endalign
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
answered Mar 26 at 6:28
Felix MarinFelix Marin
69k7110147
69k7110147
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088677%2fhow-to-prove-that-int-0-infty-ln2x-sinx2dx-frac132-sqrt-frac%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Computer algebra solves this instantly: $frac132 sqrtfracpi 2 (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
2
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12