1. ## integral question

Evaluate :

$\displaystyle \int_0^\infty \frac{x^2}{x^4+5x^2+4} dx$

$\displaystyle \int_0^\infty \frac{x^2}{(x^2+1)(x^2+4)} dx$

now I would like to extended this out into the complex plane and using the Cauchy integral to find the value, but not sure on how to do so exactly and what bounds to choose.

I figure I'll have something like:

$\displaystyle \int_C \frac{z^2}{(z^2+1)(z^2+4)} dx$

2. Do you have to use CA?.

$\displaystyle \int\frac{x^{2}}{x^{4}+5x^{2}+4}dx=\frac{4}{3}\int _{0}^{\infty}\frac{1}{x^{2}+4}dx-\frac{1}{3}\int_{0}^{\infty}\frac{1}{x^{2}+1}dx$

You should get $\displaystyle \frac{\pi}{6}$

They involve arctan

3. If it must be done with CA, here is how I would look at it.

You have:

$\displaystyle \int_{0}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}d x$

Which is correct.

Let's do $\displaystyle \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx$

Inside $\displaystyle {\sigma}(0,R)$ there are simple poles at i and 2i, with

residues $\displaystyle \frac{-1}{6i}, \;\ \frac{-4}{(-3)(4i)}=\frac{1}{3i}$,

respectively.

$\displaystyle \int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx =$

$\displaystyle 2{\pi}i\left[\frac{-1}{6i}+\frac{1}{3i}\right]=\frac{\pi}{3}$

Since we want 0 to infinity, multiply by 1/2 and get $\displaystyle \frac{\pi}{6}$

4. Originally Posted by galactus

Inside $\displaystyle {\sigma}(0,R)$ there are simple poles at i and 2i, with

residues $\displaystyle \frac{-1}{6i}, \;\ \frac{-4}{(-3)(4i)}=\frac{1}{3i}$,

respectively.
I'm just a little confused on how you got those residues...

5. Originally Posted by lllll
I'm just a little confused on how you got those residues...
The order of these poles is 1.
For a pole $\displaystyle z_0$ of order 1 of a function f, find this limit :

$\displaystyle \lim_{z \to z_0} (z-z_0)f(z)$ and it'll give the residue.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
More generally... :
A singularity $\displaystyle z_0$ is a pole of a function f if there exists a positive integer n, such that :
$\displaystyle \lim_{z \to z_0} ~ (z-z_0)^n f(z)=A \neq 0$, where A is a finite value. (n will be the order)

Then there is a direct formula giving the residue :

$\displaystyle \text{Res}_{z=z_0} ~ f(z)=\lim_{z \to z_0} ~ \frac{1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \left\{(z-z_0)^n f(z)\right\}$