- #1
kq6up
- 368
- 13
Homework Statement
Show that ##\int _{ 0 }^{ \infty }{ \frac{x^2dx}{(x^2+9)(x^2+4)^2} } =\frac{\pi}{200}##.
Homework Equations
##Res=\frac{1}{n!}\frac{d^n}{dz^n}[f(z)(z-z_0)^{n+1}]## where the order of the pole is ##n+1##.
The Attempt at a Solution
Integreading of a semicircle contour one finds poles at ##2i## and ##3i##. The second residue ##3i## was correct (I checked with mathematica). I used the ##Res=\frac{g(z)}{h^{\prime}(z)}## because it is first order, and I can just include the squared term on bottom into ##g(z)##. The first residue ##2i## is not as easy, and I need to use the equation above. ##Res=\frac{1}{1!}\frac{d}{dz}[\frac{z^2(z^2+4)^2}{(z^2+4)^2(z^2+9)}]##. This simplifies to ##\frac{d}{dz}[\frac{z^2}{z^2+9}]##. Which in turn simplifies to ##\frac{2z}{z^2+9}-\frac{2z^3}{(z^2+9)^2}##. Evaluating at ##2i## yields ##\frac{36i}{25}##. This is wrong. Mathematica yields: ##-\frac{13i}{200}##. Calculating this with the other term I got ##\frac{1}{2}2\pi i (\frac{3i}{50}-\frac{13i}{200})=\frac{\pi}{200}##. So the Mathematica term is correct, but my ##\frac{36i}{25}## is wrong. I am not sure what I did wrong here.
Any suggestions?
Thanks,.
Chris