TITLE: Integrate by the method of residue
QUESTION [6 upvotes]: 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?

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