# Integration with branch cuts and singular points

1. Nov 6, 2012

### elgen

1. The problem statement, all variables and given/known data
Prove that
$\int_0^{\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx = \frac{\pi}{2a}\frac{a^{1/\alpha}}{\sin(\pi/a)}\left(1-\cos(\pi/\alpha)\right)$ where $a>0$ and $-1<1/\alpha<1$

2. Relevant equations
It is apparent that there are two first order singular points at x=a and x=-a. Since $a^{1/\alpha}$ is a multi-valued function, i select the branch cut to start at z=0 and extend along the negative imaginary axis.

3. The attempt at a solution
The contour is the semi-circle in the upper half plane. Applying the Cauchy residue theorem, I conclude that only the arc integrations around the two simple poles contribute to the improper integral. I arrive at
$\int_{-\infty}^{+\infty}\frac{x^{1/\alpha}}{x^2-a^2} dx =-\frac{1}{2} 2\pi i \left[ Res(\frac{x^{1/\alpha}}{x^2-a^2},x=-a) + Res(\frac{x^{1/\alpha}}{x^2-a^2},x=a) \right] = -\frac{1}{2}2\pi i \left[ -\frac{1}{2a} a^{1/\alpha} e^{j\pi/\alpha} + \frac{1}{2a}a^{1/\alpha} \right]$
The result on the RHS of the above expression is not a real number.

Any comment on the possible source of error is appreciated.

elgen

2. Nov 7, 2012

### jackmell

There are some issues with that integral. For starters, just get it working for a simple case:

$$\int_0^{\infty} \frac{z^{1/2}}{z^2-4}dz$$

And it has simple poles on both the negative and positive real axis so the integral won't converge in the normal sense but will converge in the principal sense so we should write:

$$P.V. \int_0^{\infty} \frac{z^{1/2}}{z^2-4}dz$$

Now, we can still integrate along the negative real axis but will have to indent around the pole at z=-2. Also, in these cases, it's a good idea to get the numeric answer first then check your results to it. So in Mathematica:

Code (Text):

In[7]:=
NIntegrate[Sqrt[x]/(x^2 - 4), {x, 0, 2, Infinity},
Method -> "PrincipalValue"]

Out[7]=
1.1107207345395953

Ok then, does it even agree with that expression you gave above?. Can you set up the integrals to compute it symbolically now?

Edit: Just noticed maybe that's what you were implying above. What do those integrals look like then?

Last edited: Nov 7, 2012
3. Nov 7, 2012

### elgen

Thank you for the pointer. I found out that I did not express the improper integral correctly.

I took
$\int_{-\infty}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx = 2 \int_{0}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx$
which is not correct. The correct way is that
$\int_{-\infty}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx = \int_{-\infty}^{0} \frac{x^{1/\alpha}}{x^2-a^2} dx + \int_{0}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx = \lim_{R\to\infty}\int_0^R \frac{r^{1/\alpha} e^{j\theta/\alpha}}{r^2 e^{j2\theta}-a^2} dr|_{\theta=\pi} + \int_{0}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx = (1+e^{j\pi/\alpha})\int_{0}^{+\infty} \frac{x^{1/\alpha}}{x^2-a^2} dx$

Then everything falls off naturally.