Calculate $\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx$ I was asked, by a high school student in the UK, how to calculate the following integral: 
$$\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx.$$
It has been a long time since I have done any calculus and most of my immediate thoughts used mathematics that he is unlikely to have seen before.
I know that the result is $\frac{\pi^2}{12}$ but I am interested in a proof which a (good) high school student would be satisfied by. I do not mind if it goes a little beyond the A-level further maths syllabus, but I would like to avoid having to teach complex analysis or Fourier analysis just to understand this proof.
 A: Well, it is pretty much understood that
$$ \int_{0}^{+\infty}\frac{x}{\sinh x}\,dx = 2\sum_{n\geq 0}\frac{1}{(2n+1)^2}$$
so the question itself is equivalent to the Basel problem. We may notice that
$$ \frac{\pi^2}{8}=\left[\tfrac{1}{2}\arctan^2(x)\right]_{0}^{+\infty}=\int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx $$
and by Feyman's trick/Fubini's theorem the RHS can be written as
$$ \int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+a^2 x^2)(1+x^2)}\,dx\,da =\int_{0}^{1} \frac{-\log a}{1-a^2}\,da. $$
Since over $(0,1)$ we have $\frac{1}{1-a^2}=1+a^2+a^4+\ldots$ and $\int_{0}^{1}a^{2n}(-\log a)\,da = \frac{1}{(2n+1)^2}$, these manipulations prove $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$ and we are done. In a single block:
$$\boxed{\begin{eqnarray*}\int_{0}^{+\infty}\frac{x\,dx}{\sinh(\sqrt{3}x)}&\stackrel{x\mapsto z/\sqrt{3}}{=}&\frac{1}{3}\int_{0}^{+\infty}\frac{2z e^{-z}}{1-e^{-2z}}\,dz\\&=&\frac{2}{3}\sum_{n\geq 0}\int_{0}^{+\infty}z e^{-(2n+1)z}\,dz\\&=&\frac{2}{3}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\\&=&\frac{2}{3}\sum_{n\geq 0}\int_{0}^{1}a^{2n}(-\log a)\,da\\&=&\frac{1}{3}\int_{0}^{1}\frac{-\log a^2}{1-a^2}\,da\\&=&\frac{1}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{du}{(1+a^2 u)(1+u)}\,da\\&\stackrel{u\mapsto v^2}{=}&\frac{2}{3}\int_{0}^{+\infty}\int_{0}^{1}\frac{v}{(1+a^2 v^2)(1+v^2)}\,da\,dv\\&=&\frac{1}{3}\int_{0}^{+\infty}\frac{2\arctan v}{1+v^2}\,dv\\&=&\frac{1}{3}\left[\arctan^2(v)\right]_{0}^{+\infty}=\color{red}{\frac{\pi^2}{12}}.\end{eqnarray*}}$$
By this way you only have to introduce Fubini's theorem for non-negative functions, which is not a surprising result.
A: Smooth change of variables just to write it in a more "elegant" way.
$$\sqrt{3}x = y$$
The measure becomes $\text{d}y = \sqrt{3}\text{d}x$ and remember that $y = \sqrt{3}x$ which then combines with the other square root. Hence the integral becomes
$$\frac{1}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{\sinh(y)}$$
Now:
$$\sinh(y) = \frac{e^y - e^{-y}}{2}$$
And the integral is rewritten as
$$\frac{2}{3}\int_0^{+\infty} \frac{y}{e^y - e^{-y}} \ \text{d}y$$
Collect the term $e^y$ at the denominator
$$\frac{2}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{e^y(1 - e^{-2y})}$$
And make use of the geometric series for the term $\frac{1}{1 - e^{-2y}} = \sum_{k = 0}^{+\infty} e^{-2yk}$ whence
$$\frac{2}{3}\sum_{k = 0}^{+\infty} \int_0^{+\infty} y e^{-y(2k+1)}\ \text{d}y$$
Notice that the integral is now trivial:
$$\int_0^{+\infty} y e^{-y(2k+1)}\ \text{d}y = \frac{1}{(1+2k)^2}$$
What now remains is a series, which is actually trivial since:
$$\sum_{k = 0}^{+\infty} \frac{1}{(1+2k)^2} \equiv \frac{\pi^2}{8}$$
Combining with the constant and you will get eventually
$$\frac{2}{3}\frac{\pi^2}{8} = \frac{\pi^2}{12}$$
How to show the sum is $\frac{\pi}{8}$
If we write the first terms of the sum, we have:
$$1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} + \cdots + \frac{1}{n_{\text{disp}}^2}$$
This sum is indeed the sum of all the odd squares. This is a particular sum, and we can see it as the sum of ALL the reciprocal squares, minus the sum of the EVEN reciprocal squares, indeed we can write:
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \frac{1}{64} + \frac{1}{81} + \cdots + \frac{1}{n^2}$$
if we subtract from this the sum of all even reciprocal squares, we obtain exactly our sum. Translated into mathish it's like to have (calling OUR sum $S$)
$$S = \frac{1}{n^2} - \frac{1}{(2n)^2}$$
namely again: our sum is the whole sum of reciprocal squares, minus the sum of all the EVEN reciprocal squares. We can do that simple subtraction:
$$S = \frac{3n^2}{4n^4} = \frac{3}{4n^2}$$
This means that our sum is three quarters the value of the sum of all the reciprocal squares which is a well known series (also it's the Riemann Zeta vaulted in $2$):
$$\sum_{k = 1}^{\infty} \frac{1}{n^2} = \sum_{k = 1}^{+\infty} \frac{1}{(k+1)^2} = \frac{\pi^2}{6}$$
Since our sum is three quarters of that value we get:
$$S = \frac{3}{4}\cdot \frac{\pi^2}{6} = \frac{\pi^2}{8}$$
A: Noting
$$ \int\frac{1}{\sinh(x)}dx=\ln(\tanh(\frac x2))+C$$
one has
$$ \int_0^\infty\frac{x}{\sinh(\sqrt 3x)}dx=\frac13\int_0^\infty\frac{x}{\sinh(x)}dx=\frac13\int_0^\infty xd\ln(\tanh(\frac x2))=-\frac13\int_0^\infty \ln(\tanh(\frac x2))dx. $$
Let
$$ I(a)=\int_0^\infty \ln(\tanh(ax))dx=\int_0^\infty \ln\bigg(\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\bigg)dx $$
and then
$$ I'(a)=\int_0^\infty \frac{4x e^{2ax}}{e^{4ax}-1}dx=\frac{\pi^2}{8a^2}. $$
So
$$ I(\infty)-I(\frac12)=\int_1^\infty\frac{\pi^2}{8a^2}da=\frac{\pi^2}{4}$$
and hence
$$ \int_0^\infty\frac{x}{\sinh(\sqrt 3x)}dx=-\frac13\int_0^\infty \ln(\tanh(\frac x2))dx=-\frac13 I(\frac12)=\frac{\pi^2}{12}. $$
A: One may be interested in using this one which says
$$\int_0^\infty\frac{\sin(a\sqrt{3}z)}{\sinh(\sqrt{3}z)}d(\sqrt{3}z)=\frac{\pi}{2}\tanh(\frac{\pi a}{2})$$
differentiating respect to $a$ gives
$$\int_0^\infty\frac{\sqrt{3}z\cos(a\sqrt{3}z)}{\sinh(\sqrt{3}z)}d(\sqrt{3}z)=\frac{\pi^2}{4}\left(1+\tanh^2(\frac{\pi a}{2})\right)$$
now let $a=0$.
