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I am trying to evaluate:

$$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|},\textrm{ }\alpha\in\mathbb{C}\backslash(-\infty,0]$$

It's simple to see that it is convergent for such $\alpha$—but that's probably the only simple thing about it! I've been lost in trying to solve this one for several days, now. Performing a change of variables yields:

$$\int_{0}^{\infty}\frac{\sqrt{q}dx}{\sqrt{x}\sqrt{x^{2}+px+q}}$$

where $p=\frac{\cos\theta}{r}$ and $q=\frac{1}{r^{2}}$, where $\alpha=re^{i\theta}$.

Performing yet more changes of variables and integrating by parts yields (assuming I didn't screw up somewhere along the way):

$$\frac{2}{r}+\frac{2}{r}\int_{\frac{p}{2}}^{\infty}\frac{x\sqrt{x-\frac{p}{2}}}{\left(x^{2}-\Delta^{2}\right)^{3/2}}dx$$

where $\Delta=\frac{i}{2r}\sqrt{4-\cos^{2}\theta}$. This version, Mathematica is able to compute, giving the formula:

$$\int_{a}^{\infty}\frac{x\sqrt{x-a}}{\left(x^{2}-b^{2}\right)^{3/2}}dx=\frac{\textrm{sgn}\left(\textrm{arg}\left(-b^{-2}\right)\right)}{\left(a^{2}-b^{2}\right)^{1/4}}K\left(\frac{1}{2}-\frac{a}{2\sqrt{a^{2}-b^{2}}}\right)$$

where $K$ is the complete elliptic integral of the first kind. However, this formula is only valid for $a,b\in\mathbb{C}$ satisfying $\textrm{Im}\left(a\right)=0$, $\textrm{Re}\left(a\right)>0$, $\textrm{Re}\left(b^{2}\right)<0$, and satisfying either: “$\textrm{Re}\left(a\right)>\textrm{Re}\left(b\right)$ and $a>\textrm{Re}\left(b\right)$” OR “$b\notin\mathbb{R}$”.

All of these conditions are satisfied for my integral, except for the $\textrm{Re}\left(a\right)>0$ condition, which makes no sense. $a=\frac{p}{2}=\frac{\textrm{Re}\left(\alpha\right)}{\left|\alpha\right|^{2}}$, and the initial integral is valid even for $\alpha$ with $a\leq0$, as long as $\textrm{Im}\left(\alpha\right)\neq0$.

So: any ideas for how to evaluate:

$$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|}?$$

Thanks!

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  • $\begingroup$ Not that it helps much, but for positive $a$ the result seems to be $\pi/\sqrt{a}$. $\endgroup$ – mickep Apr 10 '17 at 8:02
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\alpha \in \mathbb{C}\setminus\left(-\infty,0\right].\quad}$ Lets $\ds{\alpha = \verts{\alpha}\exp\pars{\ic\phi}\quad}$ where $\ds{\quad-\pi < \phi < \pi\quad}$ and $\ds{\quad\alpha \not= 0}$.

\begin{align} &\int_{-\infty}^{\infty}{\dd x \over \verts{1 + \alpha x^{2}}} = {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {\root{\verts{\alpha}}\dd x \over \verts{\vphantom{\Large A} \verts{\alpha}x^{2} + \verts{\alpha}/\alpha}} \\[5mm] \stackrel{\root{\verts{\alpha}}x\ \mapsto\ x}{=}& {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {\dd x \over \verts{\vphantom{\Large A} x^{2} + \bar{\alpha}/\verts{\alpha}}} = {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {\dd x \over \verts{\vphantom{\Large A} x^{2} + \expo{-\ic\phi}}} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {\dd x \over \root{\pars{x^{2} + \expo{-\ic\phi}}\pars{x^{2} + \expo{\ic\phi}}}} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {\dd x \over \root{x^{4} + 2\cos\pars{\phi}x^{2} + 1}} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {1 \over \root{x^{2} + 2\cos\pars{\phi} + 1/x^{2}}}\,{\dd x \over x} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {1 \over \root{\pars{x - 1/x}^{2} + 2 + 2\cos\pars{\phi}}}\,{\dd x \over x} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\infty} {1 \over \root{\pars{x - 1/x}^{2} + 4\cos^{2}\pars{\phi/2}}}\,{\dd x \over x} \end{align} With the change of variables $\ds{t = x - {1 \over x}}$ and $\ds{x = {\root{t^{2} + 4} + t \over 2}}$: \begin{align} &\int_{-\infty}^{\infty}{\dd x \over \verts{1 + \alpha x^{2}}} = {2 \over \root{\verts{\alpha}}}\int_{-\infty}^{\infty} {\dd t \over \root{t^{2} + 4\cos^{2}\pars{\phi/2}}\root{t^{2} + 4}} \\[5mm] \stackrel{t\ =\ 2\tan\pars{\theta}}{=}\,\,\,&\ {4 \over \root{\verts{\alpha}}}\int_{0}^{\pi/2} {2\sec^{2}\pars{\theta} \over \root{4\tan^{2}\pars{\theta} + 4\cos^{2}\pars{\phi/2}}\bracks{2\sec\pars{\theta}}}\,\dd\theta \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\pi/2} {\dd\theta \over \root{\sin^{2}\pars{\theta} + \cos^{2}\pars{\phi/2}\cos^{2}\pars{\theta}}} \\[5mm] = &\ {2 \over \root{\verts{\alpha}}}\int_{0}^{\pi/2} {\dd\theta \over \root{\cos^{2}\pars{\phi/2} + \sin^{2}\pars{\phi/2}\sin^{2}\pars{\theta}}} \\[5mm] = &\ \bbx{\ds{{2 \over \root{\verts{\alpha}}} \,\mrm{K}\pars{\sin^{2}\pars{\phi \over 2}}}}\,;\qquad\alpha \not= 0\,,\quad \phi = \,\mrm{arg}\pars{\alpha}\,,\quad \phi \in \pars{-\pi,\pi} \end{align}

$\ds{\mrm{K}}$ is the Complete Elliptic Integral of the First Kind.

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  • $\begingroup$ +1) Once again, your ability to accomplish far more with far less leaves me simultaneously impressed and self-conscious. Cheers :) $\endgroup$ – David H Apr 10 '17 at 20:28
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    $\begingroup$ @DavidH Thanks. It took me so many hours. I went to sleep last night and I continued in the morning because I couldn't reproduce the '$\alpha = 1$ case'. The phase $\phi$ introduction makes it easy ( I believe so ). $\endgroup$ – Felix Marin Apr 10 '17 at 20:35
  • $\begingroup$ hey felix, nice answer (+1). is there any reason why not pulling the contour in line (3) directly up to positive infinite and integrate around the two branch cuts in the upper half of the complex plane? i guess this could save some algebra $\endgroup$ – tired Apr 10 '17 at 23:19
  • $\begingroup$ @tired Thanks. I didn't think about that. When I saw the $x^{4}$ factor I inmediately think in the '$1/x - x$-old trick'. It would yield another answer with a different point of view. Really, I got blind with $1/x - x$. Maybe, the absolute value, at the beginning, avoids to think in complex integration. $\endgroup$ – Felix Marin Apr 10 '17 at 23:46
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    $\begingroup$ @mickep It's confusing anyway. I delete the "$\arctan$" which I replaced by "$\quad\phi = \,\mathrm{arg}\left(\alpha\right)\,,\quad\phi \in \left(-\pi,\pi\right)$". Thanks for your remark. $\endgroup$ – Felix Marin Apr 11 '17 at 21:56
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Suppose $\alpha\in\mathbb{C}\setminus\left(-\infty,0\right]$, and set $\left|\alpha\right|=:\rho\in\left(0,\infty\right)\land\arg{\left(\alpha\right)}=:\theta\in\left(-\pi,\pi\right)$. Given $x\in\mathbb{R}$, we have the following expression for the modulus of the complex expression $\frac{1}{1+\alpha\,x^{2}}$ as a manifestly real function in all of its parameters:

$$\begin{align} \frac{1}{\left|1+\alpha\,x^{2}\right|} &=\frac{1}{\sqrt{\left(1+\alpha\,x^{2}\right)\left(1+\bar{\alpha}\,x^{2}\right)}}\\ &=\frac{1}{\sqrt{1+\left(\alpha+\bar{\alpha}\right)x^{2}+\alpha\bar{\alpha}\,x^{4}}}\\ &=\frac{1}{\sqrt{1+2\,\Re{\left(\alpha\right)}\,x^{2}+\left|\alpha\right|^{2}x^{4}}}\\ &=\frac{1}{\sqrt{1+2\rho\cos{\left(\theta\right)}\,x^{2}+\rho^{2}x^{4}}}.\\ \end{align}$$

As such, define the real function $J:\left(0,\infty\right)\times\left(-\pi,\pi\right)\rightarrow\mathbb{R}$ via the definite integral

$$J{\left(\rho,\theta\right)}:=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\sqrt{1+2\rho\cos{\left(\theta\right)}\,x^{2}+\rho^{2}x^{4}}}.$$

Since $J{\left(\rho,\theta\right)}$ is even in $\theta$, we may go ahead and assume WLOG that $0\le\theta<\pi$. Given real parameters $\left(\rho,\theta\right)\in\left(0,\infty\right)\times\left(0,\pi\right)$, we find

$$\begin{align} J{\left(\rho,\theta\right)} &=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\sqrt{1+2\rho\cos{\left(\theta\right)}\,x^{2}+\rho^{2}x^{4}}}\\ &=\int_{-\infty}^{\infty}\frac{\mathrm{d}y}{\sqrt{\rho}\sqrt{1+2y^{2}\cos{\left(\theta\right)}+y^{4}}};~~~\small{\left[\sqrt{\rho}\,x=y\right]}\\ &=\frac{2}{\sqrt{\rho}}\int_{0}^{\infty}\frac{\mathrm{d}y}{\sqrt{1+2y^{2}\cos{\left(\theta\right)}+y^{4}}}\\ &=\frac{2}{\sqrt{\rho}}\int_{0}^{\infty}\frac{\mathrm{d}y}{\sqrt{\left[1-2y\sin{\left(\frac{\theta}{2}\right)}+y^{2}\right]\left[1+2y\sin{\left(\frac{\theta}{2}\right)}+y^{2}\right]}}\\ &=\frac{2}{\sqrt{\rho}}\int_{0}^{\infty}\frac{\mathrm{d}y}{\sqrt{4y^{2}\left[\frac{1+y^{2}}{2y}-\sin{\left(\frac{\theta}{2}\right)}\right]\left[\frac{1+y^{2}}{2y}+\sin{\left(\frac{\theta}{2}\right)}\right]}}\\ &=\frac{2}{\sqrt{\rho}}\int_{-1}^{1}\frac{\mathrm{d}t}{\left(1-t^{2}\right)\sqrt{\left[\frac{1+t^{2}}{1-t^{2}}-\sin{\left(\frac{\theta}{2}\right)}\right]\left[\frac{1+t^{2}}{1-t^{2}}+\sin{\left(\frac{\theta}{2}\right)}\right]}};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left[1+t^{2}-\left(1-t^{2}\right)\sin{\left(\frac{\theta}{2}\right)}\right]\left[1+t^{2}+\left(1-t^{2}\right)\sin{\left(\frac{\theta}{2}\right)}\right]}}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left[1-\sin{\left(\frac{\theta}{2}\right)}+\left(1+\sin{\left(\frac{\theta}{2}\right)}\right)t^{2}\right]\left[1+\sin{\left(\frac{\theta}{2}\right)}+\left(1-\sin{\left(\frac{\theta}{2}\right)}\right)t^{2}\right]}}\\ &=\small{\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left[2\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}+2t^{2}\sin^{2}{\left(\frac{\pi}{4}+\frac{\theta}{4}\right)}\right]\left[2\sin^{2}{\left(\frac{\pi}{4}+\frac{\theta}{4}\right)}+2t^{2}\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\right]}}}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\sin^{2}{\left(\frac{\pi}{4}+\frac{\theta}{4}\right)}\left[1+\frac{t^{2}\sin^{2}{\left(\frac{\pi}{4}+\frac{\theta}{4}\right)}}{\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}\right]\left[1+\frac{t^{2}\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}{\sin^{2}{\left(\frac{\pi}{4}+\frac{\theta}{4}\right)}}\right]}}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\cos^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\left[1+\frac{t^{2}\cos^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}{\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}\right]\left[1+\frac{t^{2}\sin^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}{\cos^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}\right]}}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\cos^{2}{\left(\frac{\theta}{2}\right)}\left[1+t^{2}\cot^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\right]\left[1+t^{2}\tan^{2}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\right]}}\\ &=\frac{4}{\sqrt{\rho}}\int_{0}^{\cot{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}\frac{\sec{\left(\frac{\theta}{2}\right)}\tan{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}}{\sqrt{\left(1+u^{2}\right)\left[1+u^{2}\tan^{4}{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}\right]}}\,\mathrm{d}u;~~~\small{\left[t\cot{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}=u\right]}\\ \end{align}$$

Introducing the auxiliary parameter, $\tan{\left(\frac{\pi}{4}-\frac{\theta}{4}\right)}=:\tau\in\left(0,1\right)$, we obtain the following expression of the elliptic integral $J{\left(\rho,\theta\right)}$ in its Legendre canonical form:

$$\begin{align} J{\left(\rho,\theta\right)} &=\frac{2}{\sqrt{\rho}}\int_{0}^{\tau^{-1}}\frac{\left(1+\tau^{2}\right)}{\sqrt{\left(1+u^{2}\right)\left(1+\tau^{4}u^{2}\right)}}\,\mathrm{d}u\\ &=\frac{2\left(1+\tau^{2}\right)}{\sqrt{\rho}}\int_{0}^{\arctan{\left(\frac{1}{\tau}\right)}}\frac{\sec^{2}{\left(\varphi\right)}}{\sqrt{\left[1+\tan^{2}{\left(\varphi\right)}\right]\left[1+\tau^{4}\tan^{2}{\left(\varphi\right)}\right]}}\,\mathrm{d}\varphi;~~~\small{\left[u=\tan{\left(\varphi\right)}\right]}\\ &=\frac{2\left(1+\tau^{2}\right)}{\sqrt{\rho}}\int_{0}^{\cot^{-1}{\left(\tau\right)}}\frac{\mathrm{d}\varphi}{\sqrt{\cos^{2}{\left(\varphi\right)}+\tau^{4}\sin^{2}{\left(\varphi\right)}}}\\ &=\frac{2\left(1+\tau^{2}\right)}{\sqrt{\rho}}\int_{0}^{\cot^{-1}{\left(\tau\right)}}\frac{\mathrm{d}\varphi}{\sqrt{1-\left(1-\tau^{4}\right)\sin^{2}{\left(\varphi\right)}}}\\ &=F{\left(\cot^{-1}{\left(\tau\right)},\sqrt{1-\tau^{4}}\right)}.\blacksquare\\ \end{align}$$

As of now, I have not made any attempt to verify that the incomplete elliptic integral found in the last line above is ultimately equivalent to the complete elliptic integral produced by Mathematica, in which case we've inadvertently stumbled upon an exotic looking transformation identity that can be used to intimidate calculus students during exams, though not much else. ;)


Note: The definition for the incomplete elliptic integral of the first kind used by Wolfram Alpha and Mathematica differs from mine (which comes from DLMF):

$$F{\left(\theta,\kappa\right)}:=\int_{0}^{\theta}\frac{\mathrm{d}\varphi}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{0\le\theta\le\frac{\pi}{2}\land-1\le\kappa\le1\land\neg\left(\theta=\frac{\pi}{2}\land\kappa^{2}=1\right)}.$$


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  • $\begingroup$ (+1) Nice and much more detailed than my answer! I'm impressed that you had the motivation to write it all out. $\endgroup$ – mickep Apr 10 '17 at 19:29
  • $\begingroup$ @mickep Thank you for the kind words. Please don't be too impressed with my overly detailed answer. One of my personal projects at the moment happens to be a detailed tutorial on the Legendre reduction of general of elliptic integrals. ;) $\endgroup$ – David H Apr 10 '17 at 20:05
  • $\begingroup$ Sounds nice! Will the tutorial be available online somewhere? $\endgroup$ – mickep Apr 10 '17 at 20:33
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    $\begingroup$ @mickep When/if I get around to finishing it, at the very least I'll probably post it here: integralsandseries.prophpbb.com/forum13.html . $\endgroup$ – David H Apr 10 '17 at 21:30
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In this answer I use a variable substitution which I cannot find in the already published answers.

Say that $\alpha \neq 0$ and $\alpha = \varrho e^{i\theta}, \, -\pi <\theta< \pi$. Then $|1+\alpha x^2| = \sqrt{\varrho^2x^4 +2\varrho\cos \theta x^2+1}$ and \begin{gather*} I = \int_{-\infty}^{\infty}\dfrac{dx}{|1+\alpha x^2|} = 2\int_{0}^{\infty}\dfrac{dx}{\sqrt{\varrho^2x^4 +2\varrho\cos \theta x^2+1}} =\dfrac{2}{\sqrt{\varrho}} \int_{0}^{\infty}\dfrac{dx}{\sqrt{x^4 +2\cos \theta x^2+1}} = \\[2ex] \dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{\sqrt{x^4 +2\cos \theta x^2+1}} = \dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{\sqrt{x^4+2x^2+1-4x^2\sin^2\frac{\theta}{2}}} = \\[2ex] \dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{(x^2+1)\sqrt{1-\frac{4x^2}{(x^2+1)^2}\sin^2\frac{\theta}{2}}}.\tag{1} \end{gather*} For $0<x<1$ we put $y = \dfrac{2x}{x^2+1}, 0<y<1$. Then \begin{equation*} y(x^2+1)=2x\tag{2} \end{equation*} and \begin{equation*} x= \dfrac{1-\sqrt{1-y^2}}{y}.\tag {3} \end{equation*} From (2) we get \begin{equation*} (x^2+1)dy + 2xydx=2dx \Leftrightarrow dx= \dfrac{x^2+1}{2(1-xy)}dy = \dfrac{x^2+1}{2\sqrt{1-y^2}}dy \end{equation*} where we have used (3) in the last step. Finally we use that in (1). Thus \begin{equation*} I = \dfrac{2}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dy}{\sqrt{1-y^2}\sqrt{1-y^2\sin^2\frac{\theta}{2}}} = \dfrac{2}{\sqrt{|\alpha|}}K\left(\sin^2\frac{\theta}{2}\right). \end{equation*}

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Maybe something like this: First, we write $\alpha=a+ib$ and use the fact that the integrand is even. Thus, the integral equals $$ 2\int_0^{+\infty}\frac{1}{\sqrt{(1+ax^2)^2+(bx^2)^2}}\,dx $$ Playing a bit with that expression, we find that this equals $$ \biggl[\frac{1}{(a^2+b^2)^{1/4}} F\Bigl(2\arctan\bigl((a^2+b^2)^{1/4}x\bigr),\frac{1}{2}-\frac{a}{2\sqrt{a^2+b^2}}\Bigr)\biggr]_0^{+\infty} $$ where $F$ denotes the incomplete elliptic integral of the first kind (EllipticF in Mathematica, other conventions are also used). The limit $x=0$ gives no contribution, and, using the upper limit, one gets $$ \frac{2}{(a^2+b^2)^{1/4}}K\Bigl(\frac{1}{2}-\frac{a}{2\sqrt{a^2+b^2}}\Bigr) $$ where $K$ denotes the complete elliptic integral of the first kind, (EllipticK in Mathematica).

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Praise be! I just figured it out!

$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|}$

is the $L^{2}$ norm of $\left(1+\alpha x^{2}\right)^{-\frac{1}{2}}$. The fourier transform ($\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx$ convention) of $\left(1+\alpha x^{2}\right)^{-\frac{1}{2}}$ is $\frac{2}{\sqrt{\alpha}}K_{0}\left(\frac{2\pi\left|\xi\right|}{\sqrt{\alpha}}\right)$, where $K_{0}$ is the modified Bessel function of the second kind.

Wolfram Alpha gives

$\int_{-\infty}^{\infty}\left|K_{0}\left(\left|x\right|\right)\right|^{2}dx=\frac{\pi^{2}}{2}$

So, using Parseval's Identity, splitting the integral in half, and performing the change of variables then yields the answer:

$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|}=\frac{\pi}{\sqrt{\alpha}}$

Woo!

Wait... dammit. This doesn't take into account the fact that $\alpha$ is complex. The change of variables leads to a contour integral on a ray from 0. So there's still more work to be done.

Turns out you need to integrate the square modulus of $K_{0}$ along the ray from 0 to $\frac{\infty}{\sqrt{\alpha}}$. Any thoughts as to how to do this?

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