Random Variable wrote:

No one seems to have an evaluation of problem 44.

So here is another one.

Problem 45

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

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

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

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

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