Csdrhrt

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 $(-mathrm{i} infty, -mathrm{i}]$.

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
$$int_0^inftyfrac{1}{sqrt{x^2+1}}left(frac{1}{x-alpha}-frac{1}{x+alpha}right)dx=2alphaint_0^inftyfrac{1}{sqrt{x^2+1} (x^2-alpha^2)}dx$$
Take $x$ to be $sinh(t)$. Then
$$I=2alphaint_0^inftyfrac{1}{sinh^2(t)-alpha^2}dt.$$
Since $sinh(t)=frac{e^t-e^{-t}}{2}$, by setting $s=e^t$, we have
$$I=2alphaint_1^inftyfrac{1}{frac{1}{4}left(s-frac1sright)^2-alpha^2}frac{ds}{s}=2alphaint_0^1frac{1}{frac{1}{4}left(s-frac1sright)^2-alpha^2}frac{ds}{s}.$$
That is,
$$I=4alphaint_0^inftyfrac{s}{s^4-(4alpha^2+2)s^2+1}ds.$$
Using partial fractions,
$$I=int_0^inftyleft(frac{1}{s^2-2alpha s-1}-frac{1}{s^2+2alpha s-1}right)ds.$$
(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 $$s^2-2alpha s-1=(s-alpha-sqrt{alpha^2+1})(s-alpha+sqrt{alpha^2+1})$$ and $$s^2+2alpha s-1=(s+alpha-sqrt{alpha^2+1})(s+alpha+sqrt{alpha^2+1})$$ (using the principal branch of $sqrt{phantom{a}}$), we get
begin{align}I&=frac{1}{2sqrt{alpha^2+1}}int_0^inftyleft(frac{1}{s-alpha-sqrt{alpha^2+1}}-frac{1}{s-alpha+sqrt{alpha^2+1}}right)ds\
&phantom{aaa}-frac{1}{2sqrt{alpha^2+1}}int_0^inftyleft(frac{1}{s+alpha-sqrt{alpha^2+1}}-frac{1}{s+alpha+sqrt{alpha^2+1}}right)ds
\&=-frac{1}{2sqrt{alpha^2+1}}lnleft(frac{alpha+sqrt{alpha^2+1}}{alpha-sqrt{alpha^2+1}}right)+frac{1}{2sqrt{alpha^2+1}}lnleft(frac{alpha-sqrt{alpha^2+1}}{alpha+sqrt{alpha^2+1}}right)\&=frac{1}{sqrt{alpha^2+1}}lnleft(frac{alpha-sqrt{alpha^2+1}}{alpha+sqrt{alpha^2+1}}right)=-frac{2ln(alpha+sqrt{alpha^2+1})}{sqrt{alpha^2+1}}=-frac{2operatorname{arccosh}(-ialpha)}{sqrt{alpha^2+1}}.end{align}
The particular case $alpha=i$ yields $I=2i$.

I don't know if you'd call it elegant, but there is a closed-form antiderivative:
$$ -{frac {1}{sqrt {{alpha}^{2}+1}}ln left( {frac {-sqrt {{alpha}^{2}+1}
sqrt {{z}^{2}+1}-zalpha-1}{-z+alpha}} right) }
$$