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Problem

Compute $ \int_0^{\infty} \frac{dx}{x^a(x+2)} $, with $a \in \mathbb{R}$, using the Residue Theorem. Find the values of $a$ for which this procedure is valid.

Attempt at a solution Looking at this integrand and letting $f(z) = \frac{1}{z^a(z+2)}$, it was/is easy to show that the integral does not converge for $a \le 0$ by just using direct integration via Calculus.

For $a>0$, I noticed that there were singularities at $z=0, -2$, both of which lie on the real axis, which means we cannot take the typical approach of the Residue Theorem by considering the upper semicircle $C_R$ and taking $R \to \infty$, etc.

The next thing I attempted to consider the region bounded by the quarter half circle $C_{R'}$ with radius $R$ and the positive $Re(z)$ and $Im(z)$ axis, but also considering a quarter-circle with radius $\epsilon > 0$ about the origin in the same region since I have a singularity there. Contour integral for this region should be zero by Cauchy-Goursat, since the only possible singularity in this region would be at the origin (which I have avoided).

However, the computation becomes very messy, and this problem assumed to have no computational tools available when solving.

Am I on the right path? Or is there an easier approach to solving this question?

I also should note that, based on numerical computations on Mathematica, it seems to be valid for $0 < a < 1$.

UPDATE (10/4/2013) Problem has been resolved!

For those who are curious. Brown/Churchill's Complex Variables (8th edition) text has a discussion on an almost identical problem on p. 283. Definitely a derp moment.

1 Answers1

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If we make a branch cut along the positive real axis, and define $\log(z)$ to be real along the top of the branch cut and $\log(z)+2\pi i$ along the bottom, then $$ \lim_{R\to\infty}\int_{\gamma_R}\frac{\mathrm{d}z}{z^a(z+2)} =\int_0^\infty\frac{\mathrm{d}x}{x^a(x+2)} -\int_0^\infty\frac{e^{-2\pi ia}\,\mathrm{d}x}{x^a(x+2)}\tag{1} $$ where $\gamma_R$ is the contour $[0,R]\cup Re^{[0,2\pi i]}\cup[R,0]$. Note that the integral along $Re^{[0,2\pi i]}$ vanishes.

However, the integral along $\gamma_R$ contains the singularity at $z=-2$ where the residue is $\frac1{(-2)^a}=e^{-\pi ia}2^{-a}$. Therefore, the residue theorem and $(1)$ give us $$ 2\pi i e^{-\pi ia}2^{-a} =\left(1-e^{-2\pi ia}\right)\int_0^\infty\frac{\mathrm{d}x}{x^a(x+2)}\tag{2} $$ Solving this yields $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{x^a(x+2)} &=\frac{2\pi i e^{-\pi ia}2^{-a}}{1-e^{-2\pi ia}}\\ &=\frac{2\pi i 2^{-a}}{e^{\pi ia}-e^{-\pi ia}}\\ &=\frac{\pi\,2^{-a}}{\sin(\pi a)}\tag{3} \end{align} $$

robjohn
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  • Wow this is a beautiful answer. My complex analysis course skipped this section involving integration along branch cuts. Thank you! –  Oct 03 '13 at 22:32
  • Can someone explain a little further the first step. Where and why are we introducing the logarithm and idea of the branch cut? I can follow everything after that. – Bennett Gardiner Oct 04 '13 at 02:45
  • @Bennett $$z^{a} = e^{a \log z} = e^{a \log |z| + ia \arg z} = |z|^{a} e^{ia \arg z}$$

    If $a$ is not an integer, $z^{a}$ has multiple branches.

     – Random Variable Oct 04 '13 at 02:53
  • Thank you @RandomVariable. Makes much more sense now. – Bennett Gardiner Oct 04 '13 at 03:07
  • One more question, I see why this procedure doesn't work at $a=1$, as you lose the original integral, but I don't fully understand the restriction $0<a<1$. – Bennett Gardiner Oct 04 '13 at 03:10
  • Since the integrand behaves like $ \displaystyle\frac{1}{2x^{a}}$ near $x=0$, the integral will only converge if $0 < a< 1$. – Random Variable Oct 04 '13 at 03:25