Improper integral evaluation I'm looking for a method to evaluate the following integral:
$\displaystyle \int_0^{\infty} \left( \frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) \frac{1}{x} dx$
EDIT:
Using the link, here's what I've done so far.  Define
$F(s) = \displaystyle \int_0^{\infty} e^{-sx} \left (\frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) \frac{1}{x} dx$.  
Supposing we've shown that this integral converges nicely enough to justify differentiating underneath it, we have that 
$F'(s) = \displaystyle -\int_0^{\infty} e^{-sx} \left(\frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) dx = - \left( \int_0^{\infty}e^{-sx} \frac{e^{-x}}{2} dx + \int_0^{\infty} e^{-sx} \left(  -\frac{1}{x} + \frac{1}{e^x - 1} \right) dx \right)$.
We recognize that the first integral is the Laplace transform of $\frac{e^{-x}}{2}$ and so the integral evaluates to $\frac{1}{2(s+1)}$.
Setting $F_2(s) \displaystyle \int_0^{\infty} e^{-sx} \left(-\frac{1}{x} + \frac{1}{e^{x} - 1} \right) dx$, (and assuming a similar analysis will justify the differentiation), we find that 
$F'_2(s) = \displaystyle \int_0^{\infty} e^{-sx} - \frac{x e^{-sx}}{e^x - 1} dx = \frac{1}{s} - \int_0^{\infty} \frac{xe^{-sx}}{e^x - 1} dx $.
While I'm pretty sure I can justify the convergence of all the relevant integrals to differentiate, this last integral is not the Laplace transform of a function I'm familiar with (which doesn't mean much, because I'm not familiar with many).  Any tips on how to take care of the last integral?
 A: $$f(\lambda)=\int_0^{\infty} \frac{e^{-\lambda x}}{x}\left(\frac{1}{e^x-1}+\frac{1}{2e^x}-\frac{1}{x}\right)dx$$
$$\begin{align*}f''(\lambda)&=\int_0^{\infty} e^{-\lambda x}\left(\frac{x}{e^x-1}+\frac{x}{2e^x}-1 \right)dx\\&=\int_0^{\infty} \frac{x}{e^{\lambda x}(e^x-1)}dx+\int_0^{\infty}\frac{x}{2e^{(\lambda +1)x}}dx-\int_0^{\infty} e^{-\lambda x} dx\\&=\int_0^{\infty} \frac{x}{e^{\lambda x}(e^x-1)}dx+\frac{1}{2(\lambda+1)^2}-\frac{1}{\lambda}\end{align*}$$
Recall the Hurwitz zeta function:
$$\zeta (s,w)=\frac{1}{\Gamma (s)}\int_0^{\infty}\frac{t^{s-1}}{e^{wt}(1-e^{-t})}dt$$
So
$$f''(\lambda)=\zeta(2,\lambda+1)+\frac{1}{2(\lambda+1)^2}-\frac{1}{\lambda}$$
Now integrate back twice:
$$\int f''(\lambda)d\lambda =\frac{d}{d\lambda}\ln\Gamma (\lambda+1)-\frac{1}{2(\lambda+1)}-\ln \lambda+\mathcal{C}$$
Letting $\lambda\to\infty$ reveals that $\mathcal{C}=0$
$$\int f'(\lambda)d\lambda =\ln\Gamma (\lambda+1)+\lambda (1-\ln\lambda)-\frac{1}{2}\ln (\lambda +1)+\mathcal{C'}$$
Taking $\lambda\to\infty$ we have $\mathcal{C'}=\ln\sqrt{\dfrac{1}{2\pi}}$ (easily deduced using Stirling's approximation)
Now letting $\lambda\to 0$ gives us the result:
$$\int_0^{\infty} \frac{1}{x}\left(\frac{1}{e^x-1}+\frac{1}{2e^x}-\frac{1}{x}\right)dx=\ln\sqrt{\dfrac{1}{2\pi}}$$
