Help with $\int_{-\infty}^{\infty}\frac{dz}{(a+bz^2)^n}\:\:(a,b\geq0, n\geq1)$ (Residue Theorem)

Question

I'm stucked with that integral because the solution I get is not correct according to WolframAlpha, what I have done is the following.

If we do the semicircle contour, we have the following,

$\int_{-r}^{r}\frac{dz}{(a+bz^2)^n}+\int_{\gamma}\frac{dz}{(a+bz^2)^n}=2\pi i \sum Res(\omega,z)$

Where the second integral tends to zero when r tends to infinity and the first one tends to the integral we want to solve.

We calculate the singularities (we want the one on the upper-half plane)

$a+bz^2=0\:; z_{0,1}=\pm \:i\sqrt{\frac{a}{b}}$

As it is a pole of order n, we calculate the residue as follows:

$\frac{1}{(n-1)!}g^{(n-1)}(z_0)\:\text{where}\:g(z)=(z-z_0)^n\frac{1}{(a+bz^2)^n}=\frac{1}{(z-z_1)^n}=(z+i\sqrt{\frac{a}{b}})^{-n}$

Which we can calculate recursively and results in

$\frac{-(2n-2)!}{(n-1)!^2}\frac{1}{(2i\sqrt{\frac{a}{b}})^{2n-1}}$

So

$I=2\pi i\:\frac{-(2n-2)!}{(n-1)!^2}\frac{1}{(2i\sqrt{\frac{a}{b}})^{2n-1}}=\frac{(2n-2)!}{(n-1)!^2}\frac{\pi}{2^{2n-2}\:(\sqrt{\frac{a}{b}})^{2n-1}}$

If a set values to the parameters, I don't get the same result as WolframAlpha, but I don't know what I'm doing wrong.

Thanks :)

Answer 1

Your evalution of the residue was flawed. Note that

$$\begin{align} (z-z_0)^n \frac1{(a+bz^2)^n}&=\frac{(z-z_0)^n}{b^n (z-z_0)^n(z-z_1)^n}\\\\ &=\frac{1}{b^n(z-z_1)^n} \end{align}$$

Now, take $n-1$ derivatives with respect to $z$, then set $z=z_0$, and finally divide by $(n-1)!$.

Alternatively, we can simplify things by noting that

$$\frac{d^{n-1}}{da^{n-1}}\int_{-\infty}^\infty \frac{1}{a+bz^2}\,dz=(-1)^{n-1}(n-1)! \int_{-\infty}^\infty \frac{1}{(a+bz^2)^n}\,dz \tag1$$

Evaluting the integral on the left-hand side of $(1)$ using contour integration, we find that

$$\begin{align} \int_{-\infty}^\infty \frac1{a+bz^2}\,dz&=2\pi i \text{Res}\left(\frac1{a+bz^2}, z=i\sqrt{a/b}\right)\\\\ &=\pi/\sqrt{ab}\tag2 \end{align}$$

Now differentiate the right-hand side of $(2)$ $n-1$ times with respect to $a$ and divide by $(-1)^{n-1}(n-1)!$. Can you finish?