One integral posted came from Hermite’s integral representation of the Hurwitz zeta function. The integral is not the most difficult that I have ever evaluated, but is interesting from a pedagogical point of view. The problem is to evaluate

$$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2}$$

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

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

You’re an inspiration, RG. You sure are a Jedi at residues and contours. 🙂

Cody

Thanks Cody. I disagree about the jedi thing though: I’m more of a young padawan, still struggling. Believe me, it is progress when you know how much you don’t know.

Hey Ron – check it out, had some success with this one via a different method.

http://math.stackexchange.com/a/783004/78722

The result is related to the trigamma function. It can be calculated from 6.3.21 in http://people.math.sfu.ca/~cbm/aands/page_259.htm

This is Abramowitz & Stegun table. Any way, you did it from first principles which is always very nice.