$$I = \text{p. v.}\int_{-\infty}^{\infty}\frac{e^{\alpha x}}{e^{2x} - 1}dx$$ $$ 0<\Re(\alpha)<2$$
Use this contour (see image)
$$\lim_{\epsilon\rightarrow 0, \\ R\rightarrow\infty}{\int_{-R+i\pi}^{-R} + \int_{-R}^{-\epsilon} + \int_{C_{\epsilon^{+}}} + \int_{\epsilon}^{R} + \int_{R}^{R+i\pi} +\int_{R+i\pi}^{\epsilon+i\pi} + \int_{C_{\epsilon^{-}}} + \int_{-\epsilon+i\pi}^{-R+i\pi}}$$
Am I do right?
$$\int_{-R+i\pi}^{-R} = 0,$$ $$\int_{R}^{R+i\pi} = 0,$$ and $$\int_{C_{\epsilon^{-}}},\int_{C_{\epsilon^{+}}} = -i\pi \ \underset{z=0}{\text{Res}}\left(\frac{e^{\alpha z}}{e^{2z} - 1}\right)$$
How to do next?
For $\Re(\alpha)\in (0,2)$
$$ \text{p. v.}\int_{-\infty}^{\infty}2\frac{e^{\alpha x}}{e^{2x} - 1}dx=\lim_{y\to 0}\int_{-\infty}^{\infty}\frac{e^{\alpha (x+iy)}}{e^{2(x+iy)} - 1}dx+\int_{-\infty}^{\infty}\frac{e^{\alpha (x-iy)}}{e^{2(x-iy)} - 1}dx$$
To apply the residue theorem assume $\Im(\alpha) > 0$
$$ = i\pi \sum_{k\ge 1} e^{\alpha i\pi k}+i\pi \sum_{k\ge 0} e^{\alpha i\pi k} =i\pi \frac{1+e^{\alpha i\pi}}{1-e^{\alpha i\pi}}$$ This extends by continuity to $\alpha$ real and taking the complex conjugate to $\Im(\alpha) < 0$.