A Nasty Elliptic Integral

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newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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}}}
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$ds{alpha in mathbb{C}setminusleft(-infty,0right].quad}$ Lets
$ds{alpha = verts{alpha}exppars{icphi}quad}$ where
$ds{quad-pi < phi < piquad}$ and $ds{quadalpha not= 0}$.

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{-icphi}}}
\[5mm] = &
{2 over root{verts{alpha}}}int_{0}^{infty}
{dd x over root{pars{x^{2} + expo{-icphi}}pars{x^{2} + expo{icphi}}}}
\[5mm] = &
{2 over root{verts{alpha}}}int_{0}^{infty}
{dd x over root{x^{4} + 2cospars{phi}x^{2} + 1}}
\[5mm] = &
{2 over root{verts{alpha}}}int_{0}^{infty}
{1 over root{x^{2} + 2cospars{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 + 2cospars{phi}}},{dd x over x}
\[5mm] = &
{2 over root{verts{alpha}}}int_{0}^{infty}
{1 over root{pars{x – 1/x}^{2} + 4cos^{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} + 4cos^{2}pars{phi/2}}root{t^{2} + 4}}
\[5mm] stackrel{t = 2tanpars{theta}}{=},,,&
{4 over root{verts{alpha}}}int_{0}^{pi/2}
{2sec^{2}pars{theta} over
root{4tan^{2}pars{theta} + 4cos^{2}pars{phi/2}}bracks{2secpars{theta}}},ddtheta
\[5mm] = &
{2 over root{verts{alpha}}}int_{0}^{pi/2}
{ddtheta 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}
{ddtheta 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}}}},;qquadalpha 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.


Suppose $alphainmathbb{C}setminusleft(-infty,0right]$, and set $left|alpharight|=:rhoinleft(0,inftyright)landarg{left(alpharight)}=:thetainleft(-pi,piright)$. Given $xinmathbb{R}$, 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:

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

As such, define the real function $J:left(0,inftyright)timesleft(-pi,piright)rightarrowmathbb{R}$ via the definite integral

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

Since $J{left(rho,thetaright)}$ is even in $theta$, we may go ahead and assume WLOG that $0letheta<pi$. Given real parameters $left(rho,thetaright)inleft(0,inftyright)timesleft(0,piright)$, we find

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

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

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

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):

$$F{left(theta,kapparight)}:=int_{0}^{theta}frac{mathrm{d}varphi}{sqrt{1-kappa^{2}sin^{2}{left(varphiright)}}};~~~small{0lethetalefrac{pi}{2}land-1lekappale1landnegleft(theta=frac{pi}{2}landkappa^{2}=1right)}.$$


In this answer I use a variable substitution which I cannot find in the already published answers.

Say that $alpha neq 0$ and $alpha = varrho e^{itheta}, , -pi <theta< pi$. Then $|1+alpha x^2| = sqrt{varrho^2x^4 +2varrhocos theta x^2+1}$ and
begin{gather*}
I = int_{-infty}^{infty}dfrac{dx}{|1+alpha x^2|} = 2int_{0}^{infty}dfrac{dx}{sqrt{varrho^2x^4 +2varrhocos theta x^2+1}} =dfrac{2}{sqrt{varrho}} int_{0}^{infty}dfrac{dx}{sqrt{x^4 +2cos theta x^2+1}} = \[2ex]
dfrac{4}{sqrt{varrho}} int_{0}^{1}dfrac{dx}{sqrt{x^4 +2cos theta x^2+1}} = dfrac{4}{sqrt{varrho}} int_{0}^{1}dfrac{dx}{sqrt{x^4+2x^2+1-4x^2sin^2frac{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^2frac{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)=2xtag{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}{2sqrt{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^2sin^2frac{theta}{2}}} = dfrac{2}{sqrt{|alpha|}}Kleft(sin^2frac{theta}{2}right).
end{equation*}

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