I've been trying to find a way to integrate $\int_{-1}^{1}\frac{1}{(1+x^{2})(1-x^{2})^{1/4}}dx$ using contour integration, but I'm having a hard time coming up with a contour to use.
Since I have a branch points at $-1, 1$, and $\infty$, I need to have a branch cut that would connect all 3 of these branch points, which means I don't think I can use any sort of dog bone contour.
I also tried doing a substitution to try and make use of some sort of keyhole contour, but I was having problems with that approach as well.
Any suggestions on what to try?
Edit: This actually needs to be done using contour integration, it is not meant to be solved with other methods.
Edit: The most promising thing, so I think, is to have a contour that is a dogbone yet opens up on a line to infinity going down the positive real axis. So I have:
$* C_{1}$ Which is the top portion of the dog bone from $-1$ to $1$
$* C_{2}$ A small circle of radius $\epsilon$ around point $1$ traversed in the clockwise direction as $\epsilon$ tends to 0
$* C_{3}$ A line that extends from $1$ to $R$ as $R$ tends to $\infty$
$* C_{4}$ A circle of radius $R$ traversed in the counterclockwise direction as $R$ tends to $\infty$
$* C_{5}$ A line from $R$ to point $1$ going back towards the dog bone
$* C_{6}$ The bottom portion of the circle of radius $\epsilon$ around point $1$ traversed in the counterclockwise direction
$* C_{7}$ The bottom portion of the dog bone from $1$ to $-1$
$* C_{8}$ A circle of radius $\epsilon$ around point $-1$ traversed in the counterclockwise direction.
From here, I should be able to get everything to go to 0 except the integrals on $C_{1}$ and $C_{7}$, but instead I am stuck with $C_{3}$ and $C_{5}$ not cancelling like they should. I am not sure if this is just me having difficlty in choosing a correct branch or if I need an entirely different approach, but I feel like this is a better attempt than my previous ones.
We can express the integral in terms of elliptic integrals of the third kind $$ \Pi(a,b)=\int_0^{\pi/2}\frac{1}{(1-a\sin^2t)\sqrt{1-b\sin^2t}}\,dt. $$
First, we note that the integrand is even, and thus $$ I=2\int_0^1\frac{1}{(1+x^2)(1-x^2)^{1/4}}\,dx $$ Now, let $\sin t=(1-x^2)^{1/4}$, and you will get $$ \int_0^{\pi/2}\frac{4\sin^2(t)}{(2-\sin^4t)\sqrt{1+\sin^2t}}\,dt. $$ Now, we can do a kind of partial fraction decomposition to find that $$ \frac{4\sin^2t}{2-\sin^4t}=\frac{\sqrt{2}}{1-(1/\sqrt{2})\sin^2t}-\frac{\sqrt{2}}{1+(1/\sqrt{2})\sin^2t}. $$ Thus $$ I=\sqrt{2}\bigl(\Pi(1/\sqrt{2},-1)-\Pi(-1/\sqrt{2},-1)\bigr)\approx 1.80462. $$ The numerical value computed with Mathematica.