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How to evaluating this integral using residues where $a>0$:

$$\int _0^{\infty }\frac{x^3dx}{x^5-a^5}$$

Any help is appreciated

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    $\begingroup$ What have you attempted? To use residues you need a curve in the complex plane to integrate over, so what have you tried? $\endgroup$ – zibadawa timmy Nov 8 '13 at 3:27
  • $\begingroup$ The integral diverges as written. $\endgroup$ – Ron Gordon Nov 8 '13 at 3:31
  • $\begingroup$ Maybe the denominator is $x^5+a^5$ $\endgroup$ – Kaa1el Nov 8 '13 at 3:34
  • $\begingroup$ What I got so far: My contour goes from the origin arcs over pole at z=a going to infinity on x, then comes back to the origin at an angle $\frac{2\pi}{5}$ at the pole on line I also arc around, both arcs are in such direction as to exclude the poles from the contour so that our line integral is 0. The big arc at infinity does not contribute to anything, limit goes to 0. The path coming back at angle $\frac{2\pi}{5}$ contributes $exp(i\frac{8\pi}{5})(-I)$. The contribution of the arc at z=a I got $\frac{i\pi}{5a}$. I think that's right but having trouble getting the value of the other arc. $\endgroup$ – bart Nov 8 '13 at 3:43
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As I said, the integral posted diverges. That said, let's evaluate the following real integral:

$$\int_0^{\infty} dx \frac{x^3}{x^5+a^5}$$

To do this via residues, consider the following contour integral:

$$\oint_C dz \frac{z^3}{z^5+a^5}$$

where $C$ is a wedge of angle $2 \pi/5$ and radius $R$, with one leg along the real axis. Then the contour integral is equal to

$$\int_0^R dx \frac{x^3}{x^5+a^5} + i R \int_0^{2 \pi/5} d\theta \, e^{i \theta} \frac{R^3 e^{i 3 \theta}}{R^5 e^{i 5 \theta}+a^5} + e^{i 8 \pi/5} \int_R^0 dx \frac{x^3}{x^5+a^5}$$

As $R\to\infty$, the magnitude of the second integral vanishes as $1/R$. The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=a e^{i \pi/5}$. Thus

$$\left ( 1-e^{-i 2 \pi/5}\right) \int_0^{\infty} dx \frac{x^3}{x^5+a^5} = i 2 \pi \frac{a^3 e^{i 3 \pi/5}}{5 a^4 e^{i 4 \pi/5}} = \frac{i 2 \pi}{5 a} e^{-i \pi/5}$$

Therefore,

$$\int_0^{\infty} dx \frac{x^3}{x^5+a^5} = \frac{\pi/5}{a \sin{(\pi/5)}}$$

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  • $\begingroup$ Thanks, got everything to work out. But still a question, I see that the integral does not converge (original problem). When I convert it to the complex plane with the same contour as you described and evaluate the contribution to the integral from each part of the path, and add everything up you get a real answer. Does that tell us anything or what would that mean? $\endgroup$ – bart Nov 8 '13 at 22:34
  • $\begingroup$ @bart: not sure what you mean. Are you trying to apply the residue theorem to the original problem? What should happen is that you can compute a Cauchy principal value that avoids the pole on the axis, and you would then get some finite answer by the residue theorem. But I am not sure that's what you want. $\endgroup$ – Ron Gordon Nov 8 '13 at 22:42
  • $\begingroup$ Yes, that is what I wanted to do (apply the residue theorem) and now did and got it to work. Sorry for the confusion, I thought applying the residue theorem gives you a value for the integral. With a proper integral it does right? But with an indefinite integral it gives you a value, but is not the area under the curve. Am I understanding this right? $\endgroup$ – bart Nov 8 '13 at 23:29
  • $\begingroup$ Found the section in my book explaining the Cauchy principal value! Thanks for all the help! $\endgroup$ – bart Nov 9 '13 at 0:17
  • $\begingroup$ @bart: You're very welcome! $\endgroup$ – Ron Gordon Nov 9 '13 at 1:18

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