integration - Evaluate $int_{-infty}^infty frac{1}{sqrt{z^2 + 1}}frac{1}{z - alpha} dz$.

Evaluate

$$\int_{-\infty}^\infty \frac{1}{\sqrt{z^2 + 1}}\frac{1}{z - \alpha} dz\,.$$

What is an elegant way to evaluate this integral for Im $\alpha >0$? I imagine using residue theorem will lead to an elegant solution, such as in these related questions [1,2,3]. However I've been unable to adapt them to this line integral.

One requirement is that $\frac{1}{\sqrt{z^2 + 1}}$ be analytic in a strip around the real line $(-\infty,\infty)$. In my eyes this implies that the branch cuts can not cross the real line. For example the principal branches (parallel to the real line) or the branches $[\mathrm{i},\mathrm{i}\infty)$ and $(-\mathrm{i} \infty, -\mathrm{i}]$.

Answer

This is probably not elegant, but you can probably find a place to use the residue theorem. Let $I$ denote the integral

$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^2+1}\ (x-\alpha)}dx.$$
Then, $I$ equals
$$\int_0^\infty\frac{1}{\sqrt{x^2+1}}\left(\frac{1}{x-\alpha}-\frac{1}{x+\alpha}\right)dx=2\alpha\int_0^\infty\frac{1}{\sqrt{x^2+1}\ (x^2-\alpha^2)}dx$$

Take $x$ to be $\sinh(t)$. Then
$$I=2\alpha\int_0^\infty\frac{1}{\sinh^2(t)-\alpha^2}dt.$$
Since $\sinh(t)=\frac{e^t-e^{-t}}{2}$, by setting $s=e^t$, we have
$$I=2\alpha\int_1^\infty\frac{1}{\frac{1}{4}\left(s-\frac1s\right)^2-\alpha^2}\frac{ds}{s}=2\alpha\int_0^1\frac{1}{\frac{1}{4}\left(s-\frac1s\right)^2-\alpha^2}\frac{ds}{s}.$$
That is,
$$I=4\alpha\int_0^\infty\frac{s}{s^4-(4\alpha^2+2)s^2+1}ds.$$
Using partial fractions,
$$I=\int_0^\infty\left(\frac{1}{s^2-2\alpha s-1}-\frac{1}{s^2+2\alpha s-1}\right)ds.$$
(This is probably the place you can use the residue theorem but I am not too competent with that. Maybe you need to use a logarithm factor, and something like a keyhole contour.)

Since

$$s^2-2\alpha s-1=(s-\alpha-\sqrt{\alpha^2+1})(s-\alpha+\sqrt{\alpha^2+1})$$ and $$s^2+2\alpha s-1=(s+\alpha-\sqrt{\alpha^2+1})(s+\alpha+\sqrt{\alpha^2+1})$$ (using the principal branch of $\sqrt{\phantom{a}}$), we get
$$\begin{align}I&=\frac{1}{2\sqrt{\alpha^2+1}}\int_0^\infty\left(\frac{1}{s-\alpha-\sqrt{\alpha^2+1}}-\frac{1}{s-\alpha+\sqrt{\alpha^2+1}}\right)ds\\ &\phantom{aaa}-\frac{1}{2\sqrt{\alpha^2+1}}\int_0^\infty\left(\frac{1}{s+\alpha-\sqrt{\alpha^2+1}}-\frac{1}{s+\alpha+\sqrt{\alpha^2+1}}\right)ds \\&=-\frac{1}{2\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha+\sqrt{\alpha^2+1}}{\alpha-\sqrt{\alpha^2+1}}\right)+\frac{1}{2\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha-\sqrt{\alpha^2+1}}{\alpha+\sqrt{\alpha^2+1}}\right)\\&=\frac{1}{\sqrt{\alpha^2+1}}\ln\left(\frac{\alpha-\sqrt{\alpha^2+1}}{\alpha+\sqrt{\alpha^2+1}}\right)=-\frac{2\ln(\alpha+\sqrt{\alpha^2+1})}{\sqrt{\alpha^2+1}}=-\frac{2\operatorname{arccosh}(-i\alpha)}{\sqrt{\alpha^2+1}}.\end{align}$$
The particular case $\alpha=i$ yields $I=2i$.