I'm trying to evaluate the improper integral,
$$I(a)=\int_{0}^{\infty}\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x,~~~\text{where }a\in\mathbb{R}^+.$$
Does this integral have a simple closed form expression? And if so, how best to obtain it?
My attempt
My first idea was to integrate by parts using $f=\tan^{-1}{x}$ and $dg=\frac{\mathrm{d}x}{e^{ax}-1}$, with the hope that the resulting integral would be amenable to solution by differentiating under the integral sign. So I found the indefinite integral,
$$\int\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x=\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)-\int\left(\frac{\log{(1-e^{ax})}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x,$$
but then I realized that this would result in an imaginary boundary term for the corresponding definite integral over $[0,\infty)$ since,
$$\begin{cases}\lim_{x\to\infty}\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)=i\frac{\pi^2}{2a},\\ \lim_{x\to0}\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)=0.\end{cases}$$
I want to avoid complex variables if at all possible, so I don't know if I want to continue down this route.
Can anyone offer any hints or suggestions?
CORRECTION: As Vladimir pointed out, the correct anti-derivative is actually,
$$\int\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x=\tan^{-1}{x}\left(\frac{1}{a}\log{(e^{ax}-1)}-x\right)-\int\left(\frac{\log{(e^{ax}-1)}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x.$$
Then, since,
$$\begin{cases}\lim_{x\to\infty}\tan^{-1}{x}\left(\frac{1}{a}\log{(e^{ax}-1)}-x\right)=\frac{\pi}{2}\cdot0=0,\\ \lim_{x\to0}\tan^{-1}{x}\left(\frac{1}{a}\log{-e^{ax}-1)}-x\right)=0,\end{cases}$$
we have,
$$I(a)=-\int_{0}^{\infty}\left(\frac{\log{(e^{ax}-1)}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x.$$