# Evaluating the integral $\int_{-1}^{1} \frac{1}{(1+x^{2})(1-x^{2})^{1/4}}dx$

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.

• the searched result should be $$\frac{\sqrt{\pi } \left(\, _2F_1\left(-\frac{1}{2},1;\frac{1}{4};-1\right)-2 \, _2F_1\left(\frac{1}{2},1;\frac{1}{4};-1\right)\right) \Gamma \left(\frac{3}{4}\right)}{3 \Gamma \left(\frac{5}{4}\right)}$$ – Dr. Sonnhard Graubner Dec 4 '15 at 18:37
• Yes, I found that result as well, but I would have no idea how to produce that. I suppose it suggests some sort of series expansion (maybe?), but my knowledge on hypergeometric series is very limited. – Psymon Dec 4 '15 at 18:39
• The problem is, as you point out, is the branch point at infinity. You can see this by making a substitution $x \mapsto 1/x$ and seeing the factor of $\sqrt{x}$ in the numerator. Any detour around that branch point is sure to introduce a new integral which will end up being something like @mickep's integral. – Ron Gordon Dec 5 '15 at 19:31
• @RonGordon Yes, I'm aware of the branch point at infinity. It has definitely been the main source of my issues. Is the way to fix it just an appropriate branch or is it more complicated than that? – Psymon Dec 5 '15 at 19:44
• Doesn't matter which one - I haven't done it out to be truthful, but I have enough experience to know how it should end up. The point is that @mickep's integral provides the correct answer, and you get that by constructing the integral over the branch cut to infinity. (Now, there is a residue from the poles at $\pm i$, so that integral may actually have a different form, but in the end, then result is the same.) – Ron Gordon Dec 5 '15 at 19:56

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.