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.

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    $\begingroup$ Well, first you have $\frac{1}{\sinh(x)}=\frac{2e^{-x}}{1-e^{-2x}}=2\sum_{n=0}^\infty e^{-(2n+1)x}$. Because of that, each of the integrals in the series are straightforward: $\int_0^\infty \frac{x}{\sinh(x)} dx = 2\sum_{n=0}^\infty \frac{1}{(2n+1)^2}$ (and you can force the $\sqrt{3}$ in easily). But summing up the reciprocals of the odd squares is probably beyond the methods of regular calculus...I certainly don't know a proof that uses neither complex analysis nor Fourier analysis. You could give Euler's heuristic argument but this already really has some flavor of complex analysis to it. $\endgroup$ – Ian Sep 6 '18 at 22:08
  • $\begingroup$ You can look at the second answer at math.stackexchange.com/questions/293990/… $\endgroup$ – Biswajit Banerjee Sep 6 '18 at 22:23

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

$$\frac{1}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{\sinh(y)}$$


$$\sinh(y) = \frac{e^y - e^{-y}}{2}$$

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:

$$S = \frac{3}{4}\cdot \frac{\pi^2}{6} = \frac{\pi^2}{8}$$

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  • 2
    $\begingroup$ As I said in my comment...how can the student understand the sum of the reciprocals of the odd squares? $\endgroup$ – Ian Sep 6 '18 at 22:26
  • $\begingroup$ @Ian It's actually simple when you know how to manipulate series (this kind is still doable). I gave the answer once, in another question. Let me search for it, in order to link it here! $\endgroup$ – Von Neumann Sep 6 '18 at 22:27
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    $\begingroup$ @Ian: by studying some of the well-known proofs of the Basel problem. A short one can be derived from computing $\int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx$ through Feyman's trick, for instance. $\endgroup$ – Jack D'Aurizio Sep 6 '18 at 22:28
  • $\begingroup$ @Ian Here, search for my answer and you will find the method. Or better. I will instead add it into the answer above. math.stackexchange.com/questions/1684397/… $\endgroup$ – Von Neumann Sep 6 '18 at 22:31

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.

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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 $a=0$.

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