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I want to integrate $$\int_{0}^{\infty}\frac{1}{(1+x)^5}dx$$ by the method of residue, but I have done only problems of simple poles, but this is a problem of fifth order pole. So I am stuck in it. Also, why the value of this integral is 0 if the range is from - infinity to infinity. I think it's from cauchy principal value, but don't know why?

Roshan Shrestha
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1 Answers1

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A way to attack this via residues is to consider the following contour integral

$$\oint_C dz \frac{\log{z}}{(z+1)^5} $$

where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$. In the limit as $R \to \infty$ and $\epsilon \to 0$, the contour integral becomes equal to

$$\int_0^{\infty} dx \frac{\log{x} - (\log{x}+i 2 \pi)}{(x+1)^5} = -i 2 \pi \int_0^{\infty} \frac{dx}{(x+1)^5}$$

The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=-1$. Thus,

$$\int_0^{\infty} \frac{dx}{(x+1)^5} = -\frac1{4!} \left [\frac{d^4}{dz^4} \log{z} \right ]_{z=-1} = \frac{1 \cdot 2 \cdot 3}{4!} \frac1{(-1)^4} = \frac14$$

Ron Gordon
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  • Sir, I bet your eyes light up and start typing an answer without thinking every time you see a calculus question solvable by the residue method :D – Panglossian Oporopolist Mar 12 '15 at 11:13
  • Thanks a lot, but how to view the contour diagrammatically. – Roshan Shrestha Mar 12 '15 at 11:15
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    @RoshanShrestha: here's a picture: http://math.stackexchange.com/questions/1098670/integration-of-ln-around-a-keyhole-contour – Ron Gordon Mar 12 '15 at 11:18
  • @Ron What about cauchy principal value ? – Roshan Shrestha Mar 12 '15 at 12:02
  • @RoshanShrestha: Keep in mind your job was to apply the residue theorem to the given integral. To do that, you needed an appropriate contour and integrand from which the integral falls out, and the residue theorem provides a value for the integral. That's what I demonstated. I imagine you ask about the PV because you are having trouble grasping the material and are trying to apply the only contour you know (the semicircle in the upper half-plane) to the problem, and thus you are forced to think about PV because of the pole on the axis. Indeed, the PV over the real line is zero. (cont'd) – Ron Gordon Mar 12 '15 at 13:18
  • But that is beside the point. Sometimes, the PV is of value when performing integrals where it makes sense to introduce singularities that cancel. Other times, the PV is of interest in its own right. That is not the case here. The only pole of the integrand is outside the region of integration. As there is no symmetry about the origin in the integrand, the semicircle is not useful in finding the given integral. The keyhole contour I introduce, however, is very useful in this scenario. I'd advise studying it further. Feel free to ask questions. – Ron Gordon Mar 12 '15 at 13:21
  • @RonGordon Feel like taking a shot at a harder one? http://math.stackexchange.com/questions/848097/solving-contour-integral?rq=1 – Jerry Guern Oct 01 '15 at 15:38