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Integration using residu theorem

  • Thread starter MaxManus
  • Start date
  • #1
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Homework Statement


Solve
[tex] \int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx [/tex]
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

[tex] \frac{z^2}{(z+i)(z-i)(x+2i)(z-2i)} [/tex]

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

Answers and Replies

  • #2
CompuChip
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Homework Helper
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Would it help if I noticed that
[tex]
\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
[/tex]
?
 
  • #3
277
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Thanks, that solved the problem
[tex] \int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx [/tex]
=
[tex] 0.5*2*\pi*i*(res(i) + res(2i)) = \pi*i*(\frac{-1}{6i} +\frac{1}{3i}) = \frac{\pi}{6} [/tex]
 
  • #4
CompuChip
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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. :))
 
  • #5
277
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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.
 
  • #6
CompuChip
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Homework Helper
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Ah, so you mean that you are calculating the limit of [itex]R \to \infty[/itex] of
[tex]
\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
[/tex]

where [itex]C_R: [0, \pi] \to \mathbf{C}, t \mapsto = R e^{i t}[/itex] 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 :))
 
  • #7
277
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Yes
[tex]

\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

[/tex] = [tex] 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)} [/tex]

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
 

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