5
$\begingroup$

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.

| cite | improve this question | | | | |
$\endgroup$
  • 3
    $\begingroup$ 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
3
$\begingroup$

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:

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

| cite | improve this answer | | | | |
$\endgroup$
  • 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! $\endgroup$ – Mycroft Sep 6 '18 at 22:27
  • 1
    $\begingroup$ @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/… $\endgroup$ – Mycroft Sep 6 '18 at 22:31
7
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

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 $a=0$.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.