# residues

• May 6th 2010, 02:18 PM
ihr02
residues
Hi,

I'm working through an example of using residues to evaluate an improper integral, but I don't see how they arrive at the following initial step:

x^2 / ((x^2 + 1)(x^2 +4)) = 1/3[ (4 / (x^2 + 1) - (1 / (x^2 + 4)]

Any help would be appreciated

Thanks
• May 6th 2010, 05:04 PM
zzzoak
$\displaystyle \frac{1}{x^2+1} = \frac{1}{(x+i)(x-i)}=\frac{a}{x+i}+\frac{b}{x-i}$
• May 6th 2010, 05:33 PM
mr fantastic
Quote:

Originally Posted by ihr02
Hi,

I'm working through an example of using residues to evaluate an improper integral, but I don't see how they arrive at the following initial step:

x^2 / ((x^2 + 1)(x^2 +4)) = 1/3[ (4 / (x^2 + 1) - (1 / (x^2 + 4)]

Any help would be appreciated

Thanks

Please post the entire question. Thankyou.
• May 7th 2010, 04:11 AM
ihr02
Sorry about that. The question is using residues to evaluate the integral from 0 to infinity:

(x^2 / (x^2 + 1)(x^2 + 4))

Thanks
• May 7th 2010, 04:29 AM
mr fantastic
Quote:

Originally Posted by ihr02
Sorry about that. The question is using residues to evaluate the integral from 0 to infinity:

(x^2 / (x^2 + 1)(x^2 + 4))

Thanks

This function has simple poles at values of x such that x^2 + 1 = 0 and x^2 + 4 = 0 ....
• May 7th 2010, 05:18 AM
Opalg
Quote:

Originally Posted by ihr02
Hi,

I'm working through an example of using residues to evaluate an improper integral, but I don't see how they arrive at the following initial step:

x^2 / ((x^2 + 1)(x^2 +4)) = 1/3[ (4 / (x^2 + 1) - (1 / (x^2 + 4)]

That initial step just consists of using partial fractions to write $\displaystyle \frac{x^2}{(x^2+1)(x^2+4)} = \frac A{x^2+1} + \frac B{x^2+4}$, for suitable constants A and B. (If that looks odd, replace $\displaystyle x^2$ by a single variable, say $\displaystyle x^2=t$, to make it look more like a standard partial fractions decomposition.)

Why the worked example ahould have started in this way is a bit of a mystery. As Mr F points out, the way to solve the problem is to use the residue theorem, and it doesn't seem necessary to use partial fractions in order to find the residues.