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

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

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  • $\begingroup$ Thanks Snookie, its impressive that you deduced the right answer! $\endgroup$ – Artur Gower Dec 11 '18 at 18:34
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I don't know if you'd call it elegant, but there is a closed-form antiderivative: $$ -{\frac {1}{\sqrt {{\alpha}^{2}+1}}\ln \left( {\frac {-\sqrt {{\alpha}^{2}+1} \sqrt {{z}^{2}+1}-z\alpha-1}{-z+\alpha}} \right) } $$

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  • $\begingroup$ 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? $\endgroup$ – Artur Gower Dec 11 '18 at 18:36
  • $\begingroup$ I used Maple.... $\endgroup$ – Robert Israel Dec 11 '18 at 20:19

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