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## Integration Contest - Season 3

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Moderators: Random Variable, sos440

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 4:30 pm
galactus
Global Moderator

Posts: 902
Random Variable wrote:
No one seems to have an evaluation of problem 44.

So here is another one.

Problem 45

$\displaystyle \int_{0}^{\infty} \frac{x^{2}}{\cosh x + \cos a} \ dx = \frac{a(\pi^{2}-a^{2})}{3 \sin a} \ \ 0 < a < \pi$

Hey G:

I think I have an eval for this one.

I started with:

$\displaystyle \int_{0}^{\infty}\frac{e^{-sx}}{\cosh(x)+\cos(a)}dx=\frac{2}{\sin(a)}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin(ak)}{s+k}$....[1]

To show this, write $\displaystyle \frac{e^{-sx}}{\cosh(x)+\cos(a)}=\frac{e^{-sx}}{\sin(a)}\left(\frac{1}{1+e^{-(x+ia)}}-\frac{1}{1+e^{-(x-ia)}}\right)$

$\displaystyle =\frac{2e^{sx}}{\sin(a)}\sum_{k=0}^{\infty}(-1)^{k-1}e^{-xk}\sin(ak)$

Integrating gives the above formula in [1].

Now, take [1] and diff twice w.r.t s, then let s=0:

This gives:

$\displaystyle \int_{0}^{\infty}\frac{x^{2}}{\cosh(x)+\cos(a)}dx=\frac{4}{\sin(a)}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin(ka)}{k^{3}}$

If the sin(ak) on the right is written in terms of its exponential, then we have a trilog series.

$\displaystyle \int_{0}^{\infty}\frac{x^{2}}{\cosh(x)+\cos(a)}dx=\frac{2i}{\sin(a)}\left[Li_{3}(-e^{ia})-Li_{3}(-e^{-ia})\right]$

This is the same as $\displaystyle \frac{a(\pi^{2}-a^{2})}{3\sin(a)}$ by virtue of the identity

$\displaystyle Li_{3}(z)-Li_{3}(1/z)=-1/6log^{3}(-z)+\frac{\pi^{2}}{6}log(-z)$

$\displaystyle \frac{2i}{\sin(a)}\left(\frac{-1}{6}log^{3}(e^{ia})+\frac{\pi^{2}}{6}log(e^{-ia})\right)$

$\displaystyle =\frac{a(\pi^{2}-a^{2})}{3\sin(a)}$

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 5:00 pm
Random Variable Integration Guru

Posts: 381
You could use the Fourier series for $\displaystyle f(x) = x$, that is $\displaystyle x = 2 \sum_{k=1}^{\infty} (-1)^{n-1} \frac{\sin kx}{k}$, and integrate twice.

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 5:03 pm
galactus
Global Moderator

Posts: 902
Yeah, there's is usually many ways to go about something. Fourier can be mighty handy.

Problem 44 remains undone. Heck, G, I figured you would have went through that one like Grant through Richmond using contours :)

I dabbled with it a little, made a sub and got it down to something with a $\displaystyle x^{2}+4$ and a plain cos(x), but then got a little stuck, got too busy, and then forgot about it.

Make the sub $\displaystyle t=x-1/x$ gives:

$\displaystyle \int_{-\infty}^{\infty}\frac{1}{(t^{2}+4)(1-a\cos(t)+a^{2})}dt$

I tried the classic $\displaystyle \cos(z)=\frac{z+z^{-1}}{2}$, then kind of drifted away.

I played around a little and arrived at some residues with e^2 terms as are in the solution. May be a viable approach afterall.

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 7:17 pm
galactus
Global Moderator

Posts: 902
problem 46:

$\displaystyle \int_{0}^{\infty}\frac{\sinh(ax)\cosh(bx)}{\sinh(cx)}dx=\frac{\pi}{2c}\left(\frac{\sin(\pi a/c)}{\cos(\pi a/c)+\cos(\pi b/c)}\right)$

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 9:00 pm
Random Variable Integration Guru

Posts: 381
$\displaystyle \sinh(ax+bx) = \sinh(ax) \cosh(bx) + \cosh(ax) \sinh(bx)$

So $\displaystyle \int_{0}^{\infty} \frac{\sinh (ax) \cosh(bx)}{\sinh(cx)} \ dx = \frac{1}{2} \int_{0}^{\infty} \frac{\sinh \big((a+b)x \big) + \sinh \big((a-b )x \big)}{\sinh (cx)} \ dx$

$\displaystyle = \frac{\pi}{2c} \Big[ \tan \Big(\frac{\pi(a+b)}{2c} \Big) + \tan \Big( \frac{ \pi(a-b)}{2c} \Big) \Big]$

I tried to get my answer in that form, but I gave up after about 15 minutes.

### Re: Integration Contest - Season 3

Fri Aug 30, 2013 10:15 pm
galactus
Global Moderator

Posts: 902
I got the same thing, G. I went about it differently, though. Ultimately, I used $\displaystyle \psi(1/2+x)-\psi(1/2-x)=\pi\tan(\pi x)$

Maybe note that $\displaystyle \tan\left(\frac{\pi (a\pm b)}{2c}\right)=\frac{\sin \left(\frac{\pi (a\pm b)}{c}\right)}{1+\cos\left(\frac{\pi (a\pm b)}{c}\right)}$

It looked encouraging, but then I got tired.

### Re: Integration Contest - Season 3

Sat Aug 31, 2013 7:37 pm
galactus
Global Moderator

Posts: 902
zaidalyafey wrote:
Hi RV , nice solution . Can we prove that $\displaystyle \zeta(-2n)=0$ ?

Hey Z, I just noticed this.

I think one way to show is to use ol' Riemann's functional equation.

$\displaystyle \zeta(1-s)=2(2\pi)^{-s}\Gamma(s)\cos\left(\frac{\pi s}{2}\right)\zeta(s)$

Now, let $\displaystyle s=2n+1$

$\displaystyle \zeta(-2n)=2(2\pi)^{-(2n+1)}\Gamma(2n+1)\cos(\frac{\pi}{2}(2n+1))\zeta(2n+1)$

Note that $\displaystyle \cos(\frac{\pi}{2}(2n+1))=-\sin(\pi n)=0$, and so $\displaystyle \zeta(-2n)=0$

### Re: Integration Contest - Season 3

Sat Aug 31, 2013 7:44 pm
zaidalyafey
Global Moderator

Posts: 354
Hey C . Yes I know that but I meant using the following integral representation

Random Variable wrote:
Problem 40(b)

$\displaystyle \zeta(s) = \frac{2^{s-1}}{1-2^{1-s}} \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt$

$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Integration Contest - Season 3

Sat Aug 31, 2013 10:42 pm
Random Variable Integration Guru

Posts: 381
Problem 47

$\displaystyle \int_{0}^{1} K(k) \ dk = 2G$ where $\displaystyle K(k)$ is the complete elliptical integral of the first kind and $\displaystyle G$ is Catalan's constant

### Re: Integration Contest - Season 3

Sun Sep 01, 2013 3:10 am

Posts: 850
Location: Jaipur, India

Random Variable wrote:
Problem 47

$\displaystyle \int_{0}^{1} K(k) \ dk = 2G$ where $\displaystyle K(k)$ is the complete elliptical integral of the first kind and $\displaystyle G$ is Catalan's constant

Nice problem! These integrals are called "Moments of Elliptic Integrals".

We have

\displaystyle \begin{align*} I &=\int_0^1 \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2 \sin^2 \theta}}d\theta \ dk \\ &= \int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{\sqrt{1-k^2 \sin^2 \theta}}\ dk d\theta \\ &= \int_0^{\frac{\pi}{2}} \frac{\theta}{\sin \theta} d\theta \\ &= -\int_0^{\frac{\pi}{2}}\log \tan \frac{\theta}{2} \ d\theta \\ &= -2 \int_0^{\frac{\pi}{4}} \log \tan \theta \ d\theta \\ &= -2 \int_0^1 \frac{\log x}{1+x^2}dx \quad x=\tan \theta \\ &= -2 \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n}\log(x) \ dx \\ &= 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \\ &= 2G \end{align*}

The following closed form is also known to exist:

$\displaystyle \int_0^1 K'(k)^3 \ dk = \frac{\Gamma^8 \left(\frac{1}{4}\right)}{128\pi^2}$

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