Integration using residu theorem

[MaxManus]

Homework Statement

Solve
\int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx
works here
http://www.codecogs.com/latex/eqneditor.php

using residues

Homework Equations

The Attempt at a Solution

Can I use residues here?
The contour intergral will be a quarter of a circle in the first quadrant and lines along the positive x- and y- axis of

\frac{z^2}{(z+i)(z-i)(x+2i)(z-2i)}

But can I use Cauchy's Residue Theorem when I get divide by 0 in the contour? i and 2i is on the contour.

 

[CompuChip]

Would it help if I noticed that
<br /> \int_0^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx = \frac12 \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx <br />
?

 

[MaxManus]

Thanks, that solved the problem
\int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx
=
0.5*2*\pi*i*(res(i) + res(2i)) = \pi*i*(\frac{-1}{6i} +\frac{1}{3i}) = \frac{\pi}{6}

 

[CompuChip]

Exactly, well done.

(Of course, you would still have to prove that the identity I stated is actually true, and describe which contour you will choose.

 

[MaxManus]

Thanks

The identity is true because f(x) = f(-x) so the integral from 0 to infinity is half the integral of f from -infinity to infinity.

I'm using the contour with a half circle in the first and second quadrant with center in (0,0) and radius i equal infinity and the contour along the x-axis from -infinity to infinity.

 

[CompuChip]

Ah, so you mean that you are calculating the limit of R \to \infty of
<br /> \int_{-R}^R \frac{z^2}{(z^2+1)(z^2+4)} dz + \int_{C_R} \frac{z^2}{(z^2+1)(z^2+4)} dz<br />

where C_R: [0, \pi] \to \mathbf{C}, t \mapsto = R e^{i t} is a half-circle of radius R.

And you claim that this is equal to the original integral, that is: the second part, which you added to close the contour, does not contribute.

Provided that you all did write this down, just didn't type it out, and proved the claim, I agree :-)

(Sorry for being so annoying

 

[MaxManus]

Yes
<br /> <br /> \int_0^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx = \frac12 \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx <br /> <br /> = 0.5* \int_{C_1} \frac{z^2}{(z^2+1)(z^2+4)} dz - 0.5* \int_{C_2} \frac{z^2}{(z^2+1)(z^2+4)}

where C_1 is the closed half circle with radius infinity and C_2 is the open half circle with radius infinity.

The C_1 integral is given by Cauchy's Residue Theorem and the C_2 integral is equal to zero.

Thanks for taking the time to help me