I have trouble with this seemingly simple problem $$ K=\int_0^\infty\frac{\coth^2x-1}{(\coth x-x)^2+\frac{\pi^2}{4}}dx=\frac45.\tag{A} $$ Here is the Wolfram Alpha computation Proof.

I tried to find residue at the pole $x=\frac{\pi i}{2}$ and get $\frac{9}{5\pi i}$ (link to WA). Therefore residue theorem tell me that $K=\frac{18}{5}\neq\frac45$. I'm stuck. How to prove (A)?

  • $\begingroup$ I have the strong feeling that Glasser's master theorem is involved here. $\endgroup$ – Jack D'Aurizio Jan 22 '17 at 17:04
  • $\begingroup$ @JackD'Aurizio since $(\coth(x)-x)'=-\coth^2(x)$ the first part of the above integral is "trivial" and it remains to calulate $$I=\int_0^{\infty}\frac{1}{(x-\coth(x))^2+\pi^2/4}$$ which is superclose to a form where Glasser might be applied (see also this answer: math.stackexchange.com/questions/1015462/…) $\endgroup$ – tired Feb 8 '17 at 14:41

To answer your question, it is not correct because, while you have calculated the residue of the integrand correctly, you are assuming that the contour of integration is a semicircle. This semicircle encompasses other poles in the complex plane, which explains why your result is not right.

Evaluating a real integral via the Residue Theorem can be a tricky business. Here's a start: what contour? The semicircle seem horrible, because how do we get the other poles? Better is a closed contour that contains only the pole at $z=i \pi/2$.

Now we need to determine the form of the integrand we need in the contour integral. Remember, somehow we should get the original integral back using a direct parameterization of the contour. So, without further ado...

Consider the contour integral

$$\oint_C dz \, \tanh{\left [z-\operatorname{arctanh}{\left (z-i \frac{\pi}{2} \right )} \right ]} $$

which is equal to

$$\oint_C dz \, \frac{\displaystyle 1-z \coth{z} + i \frac{\pi}{2} \coth{z}}{\displaystyle \coth{z}-z+i \frac{\pi}{2}} $$

where $C$ is the rectangle with vertices $\pm R$ and $\pm R+i \pi$.

The contour integral is then equal to

$$\int_{-R}^R dx \frac{\displaystyle 1-x \coth{x} + i \frac{\pi}{2} \coth{x}}{\displaystyle \coth{x}-x+i \frac{\pi}{2}} + i \int_0^{\pi} dy \, \frac{\displaystyle 1-(R+i y)\coth{(R+i y)} + i \frac{\pi}{2} \coth{(R+i y)}}{\displaystyle \coth{(R+i y)} - (R+i y) + i \frac{\pi}{2}} \\ + \int_{R}^{-R} dx \frac{\displaystyle 1-x \coth{x} - i \frac{\pi}{2} \coth{x}}{\displaystyle \coth{x}-x-i \frac{\pi}{2}}\\ + i \int_{\pi}^0 dy \, \frac{\displaystyle 1-(-R+i y)\coth{(-R+i y)} + i \frac{\pi}{2} \coth{(-R+i y)}}{\displaystyle \coth{(-R+i y)} - (-R+i y) + i \frac{\pi}{2}}$$

Now consider the second and fourth integrals, i.e., those over the vertical edges of the rectangle. We consider the limit as $R \to \infty$. Note that the integrand of the second integral approaches $1$ in this limit, while the integrand of the fourth integral approaches $-1$. (Exercise for the reader.) Thus, the sum of these two integrals is $i 2 \pi$.

We can also combine the integrand of the first and third integrals to get the integrand of the integral we seek, times $i \pi$. Thus, the contour integral is equal to (after exploiting the evenness of that integrand)

$$i 2 \pi \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} + i 2 \pi$$

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i \pi/2$. Interestingly enough, this pole is not simple, but is instead a third-order pole. Given the integrand, I find it easier to compute the Laurent series directly and use that to find the residue. However, I will simply state the result as follows:

$$\operatorname*{Res}_{z=i \pi/2} \frac{\displaystyle 1-z \coth{z} + i \frac{\pi}{2} \coth{z}}{\displaystyle \coth{z}-z+i \frac{\pi}{2}} = \frac{9}{5} $$

so that

$$i 2 \pi \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} + i 2 \pi = i 2 \pi \frac{9}{5} $$


$$ \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} = \frac{4}{5} $$


Let's take a look at that residue calculation. For this, it helps to know that

$$\coth{\left ( z + i \frac{\pi}{2} \right )} = \tanh{z}$$


$$\tanh{z} = z - \frac13 z^3 + \frac{2}{15} z^5 + O \left ( z^7 \right )$$

so that the integrand looks like, in the neighborhood of $z=i \pi/2$,

$$ \frac{1-z \tanh{z}}{\tanh{z}-z} $$

Now we can find the Laurent expansion of this function about $z=0$, which looks like

$$-\frac{3}{z^3} \frac{\displaystyle 1-z^2 +O \left ( z^4 \right )}{\displaystyle 1-\frac{2}{5} z^2+O \left ( z^4 \right )} = -\frac{3}{z^3} \left [1-z^2 +O \left ( z^4 \right ) \right ] \left [1+\frac{2}{5} z^2+O \left ( z^4 \right ) \right ]$$

The residue is the coefficient of $z^2$ in the numerator, which may simply be read off as $9/5$ as stated above.


How do we show that $z=i \pi/2$ is the only pole inside the rectangle? We can use Rouche's theorem. For example, we would just need to show that, on the rectangle,

$$\left | \coth{z}-z \pm i \frac{\pi}{2} \right | \gt |\coth{z} | $$

On the horizontal sides of the rectangle, we see that

$$\left | \coth{x}-x \pm i \frac{\pi}{2} \right |^2 - |\coth{x} |^2 = \frac{\pi^2}{4} - \left (2 x \coth{x}-x^2 \right ) $$

which is indeed $\gt 0$ for all $x \in \mathbb{R} $.

On the vertical sides of the rectangle, the inequality is obviously satisfied because $|\coth{(R + i y)}|$ approaches $1$ as $R \to \pm \infty$.

Therefore, by Rouche's theorem, the denominator of the integrand has the same number of zeroes inside the rectangle as $\coth{z}$, which is just the one at $z=i \pi/2$ and no others. Thus, the only pole inside the rectangle is at $z=i \pi/2$.

  • $\begingroup$ Good work! Thanks! $\endgroup$ – Tyrell Feb 6 '17 at 19:34
  • $\begingroup$ An outstanding work @Ron. $\endgroup$ – user348832 Feb 11 '17 at 6:43
  • $\begingroup$ sorry to bother @Ron does the above method apply the same for $\int_{0}^{\infty}{coth^2x-1\over (cothx+x)^2+\pi^2/4}={1\over 2}$? only a sign changed to to $\color{red}+x$ $\endgroup$ – user348832 Feb 11 '17 at 6:55
  • $\begingroup$ @TheFlAsH: it should. $\endgroup$ – Ron Gordon Feb 15 '17 at 13:27

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