1. ## Dumbbell Contour Integral

(First off I hope I am posting in the right location)

So I am pretty new to complex analysis, but I seem to have a handle on most contour integrals that can be solved by direct use of the residue theorem and I am getting stuck dealing with Branch cuts.

Particularly I am dealing with what I think is a frustrating integral.

Using contour integration I am suppose to deal with
$\int \!{\frac {1}{ \left( {a}^{2}+{x}^{2} \right) \sqrt {1-{x}^{2}}}}{
dx}
$

from -1 to 1

I figured out that I want to use a branch cut from -1 to 1 and a contour that goes counter-clockwise around the two branch points at -1 and 1 in the shape of a dumbbell (or dogbone) with the semi-circle ends near -1 and 1.

Any help would be great.

Thanks

2. Okay, and that integral is the sum of the residues at x= 1 and x= -1. Have you found them? After you know the value of the contour integral, look at the circles on the ends. What is the limit of those integrals as the radius goes to 0? If you subtract that off, what is left is the integrals along the lines- that may be a problem because you are integrating from -1 to 1 and from 1 to -1.

3. There are no residues at the branch points since these are non-isolated singularities. This is what I'd do:

You have:

$f(z)=\frac{1}{(z-ia)(z+ia)\sqrt{1-z^2}}$

That's a multi-function (with multi-residues) consisting of two sheets each of which is analytic outside the branch points and poles of the function. Now, chose one of those sheets and consider the dog-bone contour AND two contour around the poles, AND another large contour with $|z|>a$ (or max of (1,a)) over that sheet. Now, all these contours are over a totally analytic function and so the integral over the contour $|z|>a$ is then equal to the sum of the integrals over the others right? That's:

$\mathop\int\limits_{|z|>a} f_1(z)dz=\mathop\int\limits_{\text{bone}} f_1(z)dz+\mathop\int\limits_{z=ai+\rho e^{it}} f_1(z)dz+\mathop\int\limits_{z=-ai+\rho e^{it}} f_1(z)dz$

Where I'm using $f_1(z)$ to designate my analytic sheet. Now, you can calculate the integral over $|z|>a$ using the residue at infinity. You can calculate the integral over the bone by remembering it's across a branch-cut, so it's value below the real axis is minus it's value above and the contours around the branch-points as the radius goes to zero you can figure out. Now, as far as the integrals around each pole, you can calculate those using the CORRECT value of the residue based on the "determination" of the square root at each pole. So if we choose the sheet so that $\sqrt{1-z^2}\equiv \sqrt{1-z^2}$ above the real axis, than use that value in the calculation of the residue there. But below the real axis, $\sqrt{1-z^2}\equiv -\sqrt{1-z^2}$.

4. Ok so I understand the dogbone part (if I find the residues and compute the dogbone contour's associated with the integral) I get $2*Pi/(a*sqrt(1+a^2)))$, but I don't quite understand how to do the rest.

If I plug the integral into wolfram alpha I end up with :

$integral_(-1)^11/((a^2+x^2) sqrt(1-x^2)) dx = (pi sqrt(1/a^2))/sqrt(a^2+1)$

I am pretty confident that i am missing something

5. You understand the residue at infinity right? Look at the plot below which shows what we'e integrating over except I'm not drawing the other contours cus' it's tedious. Then:

$\mathop\oint\limits_{\text{yellow}} f_1(z)dz=\int_{1}^{-1} f(x)dx+\int_{-1}^1 (-f(x))dx+\text{Residue part}$

So that taking the residue at infinity and simplifying:

$2\pi i \mathop\text{Res}_{z=\infty} f_1(z)=-2\int_{-1}^1 f(x)dx+\text{Residue part}$

$2\pi i\mathop\text{Res}_{z=0} \frac{1}{z^2} f_1(1/z)=-2\int_{-1}^1 f(x)dx+\text{Residue part}$

And it just so happens that the residue at infinity is the same for either sheet so that you can just substitute $z\to 1/z$ into f(z) and show that the residue at infinity is in fact zero since the numerator of $1/z^2 f(1/z)$ has a term greater than $z^2$ right?

Also, I'm not an expert in this so I may have a few loose ends as far as rigorously defining the sheets we're using.