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

• Note that $\displaystyle\,\sqrt{\,{z^{2} + 1}\,}\,$ has branch-cuts at $\displaystyle \pm\mathrm{i}$. – Felix Marin Dec 10 '18 at 18:40

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 $$I=2i$$.
I don't know if you'd call it elegant, but there is a closed-form antiderivative: $$-{\frac {1}{\sqrt {{\alpha}^{2}+1}}\ln \left( {\frac {-\sqrt {{\alpha}^{2}+1} \sqrt {{z}^{2}+1}-z\alpha-1}{-z+\alpha}} \right) }$$
• Thanks Robert, this anti-derivative is what has allowed me to split the function $1/\sqrt{z^2 +1}$ into a Wiener-Hopf decomposition. How did you find it? – Artur Gower Dec 11 '18 at 18:36