Evaluating $\int_{-\infty}^{\infty} \frac{z\ln(1+e^z) }{(1+z^2)^2}dz = \frac{\pi}{4}$ - how to handle branch cuts? My aim is to evaluate the integral
$$I = \int_{-\infty}^{\infty} \frac{z\ln(1+e^z) }{(1+z^2)^2}dz = \frac{\pi}{4}$$
directly with contour integration. How can I do this? In particular, how should I handle the many branch points associated with $z=(2n+1)\pi i$ for each integer $n$?

On a naive attempt, I equated $I$ with a semi-circle contour integral of the same function in the upper half plane (the function should be decreasing in $|z|$ fast enough so that the arc won't contribute), which would pick up the residue from $z=i$. Naively, that would be $2 \pi i \frac{d}{dz} \frac{z\ln(1+e^z)}{(1+z^2)^2}$ evaluated at $z=i$, which would yield $\frac{\pi}{2(1+e^{-i})}$.
Unfortunately, there shouldn't be an imaginary part, but the real part looks ok. Normally with such occurrences, I would just drop the imaginary part and continue. However, I want to understand what's going wrong, and how to draw an appropriate contour or identify additional contributions.
 A: Here is a completed attempt with one choice of branch cuts.
I will define $\ln(z)$ to have a branch cut on the positive real axis.
Given that $\ln(1+e^z)$ has branch points at $z=(2n+1)\pi i$, and $1+e^z$ has positive real part for $\Re(z)<0$ with sign of $\Im(z)$ changing as one passes through $\Im(z)=(2n+1)\pi$, then $\ln(1+e^z)$ has branch cuts going horizontally to the left from the branch points.
I will thus take the following contour:

I found the contribution from the double pole at $z=i$ in the OP, and the arc is again negligible, yielding
$$\frac{\pi}{2(1+e^{-i})}= I + \sum_{n=0}^\infty \int_{-\infty}^0 (2\pi i) \frac{x+(2n+1)\pi i}{(1+(x+(2n+1)\pi i)^2)^2}$$
The left hand-side is the contribution from the double pole, and the rightmost sum is, for a given $n$, from the rightgoing + leftgoing horizontal lines following the branch cut about the branch point at $(2n+1)\pi i$. It is straightforward to integrate, using $\int dx \frac{x}{(1+x^2)^2} = -\frac{1}{2(1+x^2)}$.
$$\frac{\pi}{2(1+e^{-i})}= I + \sum_{n=0}^\infty (2\pi i) \frac{-1}{2(1+((2n+1)\pi i)^2)}$$
Summing (via $\sum_{n=0}^\infty \frac{1}{((2n+1)\pi)^2-a^2} = \frac{\pi}{4a} \tan(\frac{ \pi a}{2})$) and rearranging, one finds
$$I = \frac{\pi}{2(1+e^{-i})}-\frac{\pi i}{4}\tan(1/2)$$
After a little trig, this simplifies to
$$I = \frac{\pi}{4}$$
which completes this problem.
A: Too long for a comment
If I understand correctly your program of investigation.
If you consider your initial integral, all branches of logarithm should be delt carefully. One of the possible way is to choose the contour closed in the upper half-plane with the cut from $z=\pi i$ to $i\infty$ - which absorbs all branch points in the upper half-plane. While integrating along right and left bank of the cut you should take into consideration the phases (which are different on these banks). Integral along big quarter-circles vanish. Integrals along the closed contour gives you the residual in the single pole (at $z=i$).
I'm not sure whether this program is exactly you want, but is one of the possible ways...

