Using the residue theorem to evaluate a real integral.

  • Thread starter jrp131191
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  • #1
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Hi, I am trying to evaluate the integral from 0 to infinity of x^2/[(x^2+9)(x^2+4)^2 dx
(really really sorry, I don't know how to use latex yet but as of today I will begin learning!)

I chose to evaluate it around a semi-circle of radius R, R>3 (y>0 quadrants) and have found that there are poles at +/-3i and +/-2i. Though only two poles, 3i and 2i lie inside my contour.

I've calculated the residues and confirmed that they are indeed correct with wolfram. I suspect I'm doing something wrong with my region of integration since the integration limits are from 0-->infinity I may have chosen a stupid contour?

My final answer is pi/100 whereas wolfram calculates pi/200..
 

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  • #2
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Ok I think I figured it out.. I'm integrating an even function so if my integral is I(z) then I(z)=(1/2)J(z) where J(z) is the integral from -inf to inf... and I've actually integrated J(z) correct? I've just kind of read this off an example I've seen so If anyone can explain to me what was wrong it would be much appreciated.
 
  • #3
Stephen Tashi
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The latex would be:

I am trying to evaluate the integral [itex] \int_0^\infty \frac{x^2}{(x^2+9)(x^2 + 4)} dx [/itex]

I evaluated it around a semi-circle of radius [itex] R, R>3 [/itex] in the [itex]y>0 [/itex] quadrants. There are poles at [itex] \pm -3i [/itex] and [itex] \pm+/-2i [/itex]. Onlly two poles, [itex] 3i [/itex] and [itex] 2i [/itex] lie inside my contour.

I calculated the residues and confirmed that they are correct with Wolfram. My integration limits are from [itex] 0 [/itex] to [itex] \infty [/itex]. Did I chosen a stupid contour?

My final answer is [itex] \frac{\pi}{100} [/itex] whereas Wolfram calculates [itex] \frac{\pi}{200} [/itex]
 

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