A Nasty Elliptic Integral

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!

Solution 1:

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\alpha \in \mathbb{C}\setminus\left(-\infty,0\right].\quad}$ 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.

Solution 2:

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):

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