Can I use the residue calculus here? I know from an alternative to the residue calculus that $$I=\int_{-\infty}^{\infty}\frac{e^x}{e^{e^x}+1}dx = \log 2.$$
However, I see no reason I cannot apply the residue calculus here.
My attempt:
$$I+iR\int_0^\pi \frac{e^{R e^{i \theta}}}{e^{e^{Re^{i\theta}}}+1}d\theta=2\pi i\sum\text{residues},$$ and it seems that the semi-circular integral $\to 0$ as $R\to\infty$. Hence, we are left with the problem of summing the residues, which I think might be where the problem lies:
I've worked through the algebra to obtain poles at
$$z_{k,\ell}=\log\left|\pi(2k+1)\right| + i\frac{\pi}{2}(2\ell+1).$$
Hence,
$$I = 2\pi i\sum_{\substack{k\in\mathbb{Z}\\ \ell\geq 0}}\text{Res}\left(\frac{e^z}{e^{e^z}+1};z_{k,\ell}\right).$$
But Mathematica evaluates all residues to $-1$, which gives a dvergent series.
What am I doing wrong? My thoughts are that the order of the function in the denominator of the integrand is where the problem lies. However, I'm pretty certain the number of poles is countable since it seems pretty clear there's a 1-1 bijection with $\mathbb{Q}$.
EDIT
As Claude Leibovici points out, the anti-derivative has elementary function form
$$e^x - \log(1 + e^{e^x}),$$ but I just wanted to work out the integral in as many ways as I hope possible.
 A: I think that the problem is due to the infinite collection of residues. Usually, one has a finite collection of poles (when $R$ is large enough) so that the integral along the upper arc is easy to estimate away. However, in this case, as $R$ gets larger you keep collecting more and more poles, which is very different.
It's also not clear to me that the expression inside the integral goes to zero for large values of $R$! If you look at the denominator, it goes roughly as
$$
\begin{align}
\left|e^{e^{Re^{i\theta}}} + 1\right| &\approx \left|e^{e^{Re^{i\theta}}}\right| \\
&=\left|e^{e^{R\cos\theta}e^{iR\sin\theta}}\right| \\
&=\left|e^{e^{R\cos\theta}(\cos(R\sin\theta)+i\sin(r\sin\theta))}\right| \\
&=\left|e^{e^{R\cos\theta}\cos(R\sin\theta)}\right|
\end{align}
$$
However, for $R \to \infty$, $\cos(R\sin\theta)$ oscillates wildly, and so it doesn't seem obvious to me that this term gets large for large $R$.
It might be possible to choose a different path: make a box that runs up, then across, then back down in straight lines; this way you could more easily avoid the poles, although the computation may be pretty hard nonetheless.
