# Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$

The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique?

$$\int_{0}^{\infty} {x \over \left({\rm e}^{x} - 1\right)\left[x^{2} + \left(2\pi\right)^{2}\right]^{2}} \,{\rm d}x ={1 \over 96} - {3 \over 32\pi^{2}}.$$

This integral may be done via the residue theorem, by considering the integral

$$\oint_C dz \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2}$$

where $C$ is a keyhole contour about the positive real axis, having an outer circle of radius $R$ and inner circle of radius $\epsilon$, each centered about the origin. One may show that the integrals about each of the circular arcs vanish as $R\to\infty$ and $\epsilon \to 0$. In this case, then, by the residue theorem, we have

$$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2} = -\sum_{k \ne 0, k \in \mathbb{Z}} \operatorname*{Res}_{z=i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2}$$

When $k \ne 1$, the poles are simple. Noting that

$$\lim_{z\to i 2 \pi k} \frac{z-i 2 \pi k}{e^z-1}=1$$

we have that, when $k \gt 1$

$$-\operatorname*{Res}_{z=i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} = \frac{k}{16 \pi^2 (k^2-1)^2}- i \frac{k \log{(2 \pi k)}}{8 \pi^3 (k^2-1)^2}$$

When $k \lt 0$, however, we must be careful with the argument of $-i$ as defined by the keyhole contour we are using; in this case, $\arg{(-i)} = 3 \pi/2$ and not $-\pi/2$, and therefore, when $k \lt -1$

$$-\operatorname*{Res}_{z=-i 2 \pi k} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} = -\frac{3 k}{16 \pi^2 (k^2-1)^2}+ i \frac{k \log{(2 \pi k)}}{8 \pi^3 (k^2-1)^2}$$

Thus the sum of the residues when $k \ne \pm 1$ is

$$-\frac1{8 \pi^2} \sum_{k=2}^{\infty} \frac{k}{(k^2-1)^2}$$

The sum may be evaluated by partial fractions:

$$\sum_{k=2}^{\infty} \frac{k}{(k^2-1)^2} = \sum_{k=1}^{\infty} \frac{k+1}{((k+1)^2-1)^2} = \sum_{k=1}^{\infty}\frac14 \left [\frac1{k^2}-\frac1{(k+2)^2} \right ] = \frac{5}{16}$$

Putting this together gives the sum of the residues for $|k|\gt 1$ as $-5/(128 \pi^2)$.

For $|k|=1$, however, we have a triple pole. The calculation in this case is far uglier and I will spare you the details:

\begin{align}-\operatorname*{Res}_{z=i 2 \pi} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} &= -\frac12 \left [\frac{d^2}{dz^2}\frac{z (z-i 2 \pi) \log{z}}{(e^z-1)(z+i 2 \pi)^2} \right ]_{z=i 2 \pi}\\ &= -\frac1{192}-\frac{9}{256 \pi^2} -i \frac{(3+4 \pi^2) \log{(2 \pi)}}{384 \pi^3} \end{align}

Similarly, and keeping in mind the argument of $-i$ as above, we have

\begin{align}-\operatorname*{Res}_{z=-i 2 \pi} \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2} &= -\frac12 \left [\frac{d^2}{dz^2}\frac{z (z+i 2 \pi) \log{z}}{(e^z-1)(z-i 2 \pi)^2} \right ]_{z=-i 2 \pi}\\ &= \frac{3}{192}-\frac{5}{256 \pi^2} +i \frac{(3+4 \pi^2) \log{(2 \pi)}}{384 \pi^3} \end{align}

We may put this all together, and we finally have for the integral

$$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2} = \frac1{96}-\frac{7}{128 \pi^2} - \frac{5}{128 \pi^2} = \frac1{96}-\frac{3}{32 \pi^2}$$

as was to be shown.

• Just wanted to note that it was enough to find the integral $$F(a)=\int^\infty_0\frac{1}{(e^x-1)(x^2+a^2)}\, dx$$ Then differentiate under the integral sign to get the our integral. – Zaid Alyafeai Jan 14 '14 at 2:57
• @ZaidAlyafeai: lucky you caught me in a generous mood, as I left the simple way to solve the problem to you. Good luck! – Ron Gordon Jan 14 '14 at 3:17
• @ZaidAlyafeai: introducing the log is a standard method of evaluating some real integrals via residues. You would likely not think of it unless you had some experience there. – Ron Gordon Jan 14 '14 at 3:26
• @RonGordon - I think Zaid meant $$F(a) = \int_0^\infty\frac{x}{(e^x-1)(x^2+a^2)}dx.$$ Differentiating this with respect to $a$ will, after dividing by $-2a$, give the desired integral. – Pixel Jan 14 '14 at 15:28
• @Zaid It will be simpler in the sense there won't be a triple pole, only a double pole! – Pixel Jan 14 '14 at 15:45

Binet's second Log Gamma integral formula is $$\ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t$$ Differentiating twice w.r.t. $z$ gives $$\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 4\int_0^{\infty} \frac{zt}{(z^2+t^2)^2(e^{2\pi t}-1)} \ \mathrm{d}t.$$ Let $t \mapsto x/2\pi$, then $$\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + \int_0^{\infty} \frac{zx}{\pi^2(z^2+x^2/4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x,$$ $$\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 16\pi^2\int_0^{\infty} \frac{zx}{(4\pi^2z^2+x^2)^2(e^{x}-1)} \ \mathrm{d}x.$$ Evaluate at $z=1$, $${\pi^2 \over 6} = \frac{3}{2}+16\pi^2\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x,$$ or, rearranging - $$\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x = \frac{1}{96} - \frac{3}{32\pi^2}.$$

• that's great, thanks. – Pixel May 6 '14 at 8:41
• Thanks, it's not very generalisable like Ron's, and it required the formula to begin with, but I'm still proud of this one. Great question! – Bennett Gardiner May 6 '14 at 12:50