The Laurent series about $0$ of $1/{z(e^z-1)}$ is
$$\frac{1}{z(e^z-1)}=\frac{1}{z^2}-\frac{1}{2z}+\sum_{n=2}^\infty\frac{B_n}{n!}z^{n-2}$$
When we integrate this we have
$$\int\frac{1}{z(e^z-1)}dz=-\frac{1}{z}-\frac{\ln{|z|}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}z^{n-1}+C$$
Something interesting happens when we subtract the expansion from the definite integral form. It equals a constant.
$$C_0=\int_\infty^x\frac{1}{z(e^z-1)}dz-\left(-\frac{1}{x}-\frac{\ln{|x|}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right)$$
Unfortunately the series is only defined on $0\lt x\lt 2\pi$.
What is the value of the constant $C_0$?
I thought about seeing if I can do the expansion about some other point $a$, so here's what I did
$$\begin{align}\frac{1}{z(e^z-1)}&=\frac{1}{a(z/a)(e^{a(z/a)}-1)}\\&=\sum_{n=0}^\infty\frac{B_n}{n!}a^{n-2}\left(\frac{z}{a}\right)^{n-2}\\&=\sum_{n=0}^\infty\frac{B_n}{n!}a^{n-k-2}\sum_{k=0}^\infty{n-2\choose{k}}\left(z-a\right)^k\\&=\sum_{k=0}^\infty c_k\left(z-a\right)^k\end{align}$$
Where
$$c_k=\sum_{n=0}^\infty\frac{B_n}{n!}{n-2\choose{k}}a^{n-k-2}$$
Using the Euler-Maclaurin summation formula with $f(x)=\frac{(-1)^k}{k!}\Gamma(k+1,-x)$ we get
$$c_k=\frac{\left(-1\right)^k}{a^{k+1}k!}\sum_{n=1}^\infty \Gamma\left(k+1,na\right)$$
So by integration we have
$$\int\frac{1}{z(e^z-1)}dz=\sum_{k=0}^\infty \frac{c_k}{k+1}\left(z-a\right)^{k+1}+C$$
Then
$$\int_\infty^x\frac{1}{z(e^z-1)}dz=\lim\limits_{a\to\infty}\sum_{k=0}^\infty \frac{c_k}{k+1}\left(x-a\right)^{k+1}$$
Then we can try subtracting the original series expression that is defined on $(0,2\pi)$ from this expression that is defined on $(0,\infty)$ for some $x\in (0,2\pi)$.
