The problem in the book is

Compute $\int_0^\infty \frac{dx}{x^3+8}$.

I set up the keyhole contour, apply the residue theorem, and go through the tedious algebra. I get stuck in doing so, but looking at the solution, I see the following remark:

Note then that $\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$.

Here, $z_k$ represents one of the poles of the function $\frac{\log z}{z^3+8}$. What I want to know is how the author makes this claim.

I am trying to compute the residues using the simple formula

$$\textrm{Res}(f(z); z_k) = \frac{A(z_k)}{B'(z_k)}$$

when $z_k$ is a simple pole. In this case, letting $B(z) = z^3+8$, I have the following:

$$z_k = -2, 1\pm\sqrt{3}i.$$

This yields the following residues

$$\begin{align*} \textrm{Res}\left(\frac{\log z}{z^3+8}; z=-2\right) &= \frac{1}{12}\left(\log 2 + i\pi\right), \\ \textrm{Res}\left(\frac{\log z}{z^3+8}; z=1+\sqrt{3}i\right) &= \frac{2}{12(-1+\sqrt{3}i)}\left(\log 2 + \frac{i\pi}{3}\right), \\ \textrm{Res}\left(\frac{\log z}{z^3+8}; z=1-\sqrt{3}i\right) &= \frac{2}{12(-1-\sqrt{3}i)}\left(\log 2 - \frac{i\pi}{3}\right). \end{align*} $$

However, when I sum the results, I get a non-zero real and imaginary part. What am I doing wrong?

  • $\begingroup$ It might be that you are not working within a consistent branch of $\log{z}$. You define your arguments of the poles at $\pm \pi/3$, but I think the log is expecting $\pi/3$ and $5 \pi/3$. $\endgroup$ – Ron Gordon Mar 19 '13 at 18:52

Things are a lot easier if you choose a contour which is a circular wedge shape encompassing a single pole. I chose a wedge angle of $2 \pi/3$ and used the residue theorem with just the one pole at $z=\pi/3$ and got the correct result, $\pi/(6 \sqrt{3})$. Let me explain.


$$\oint_C \frac{dz}{z^3+8}$$

where $C$ is the contour that goes from $[0,R]$ along the real axis, then along a circular arc of angle $2 \pi/3$ at radius $R$, then along the line $z=t e^{i 2 \pi/3}$ from $t=R$ to $t=0$. The integral along the circular arc vanishes as $2 \pi/(3 R^2)$ as $R \rightarrow \infty$. So we are left with

$$\oint_C \frac{dz}{z^3+8} = e^{i 2 \pi/3} \int_{-\infty}^0 \frac{dt}{t^3+8} + \int_0^{\infty} \frac{dx}{x^3+8}$$

This is equal to i 2 \pi times the residue at the only pole within $C$, namely that at $z=2 e^{i \pi/3}$. I will let you work out the algebra inevaluating this rsidue; the bottom line is that

$$\oint_C \frac{dz}{z^3+8} = (1-e^{i 2 \pi/3}) \int_0^{\infty} \frac{dx}{x^3+8} = \frac{\pi}{6} e^{-i \pi/6}$$

Multiply both sides by $e^{i \pi/6}$, noting that $e^{i \pi/6} - e^{i 5 \pi/6} = \sqrt{3}$. Then it follows that

$$\int_0^{\infty} \frac{dx}{x^3+8} = \frac{\pi}{6 \sqrt{3}}$$

as was to be shown.

  • $\begingroup$ BTW, credit to @robjohn for introducing me to this technique. $\endgroup$ – Ron Gordon Mar 19 '13 at 18:50
  • $\begingroup$ Thanks -- I think I found my error when doing it using a keyhole contour (just a mess of minus signs), but I'm still curious how the author makes his first remark. Where does the 24 in the denominator come from?? $\endgroup$ – Emily Mar 19 '13 at 18:52
  • $\begingroup$ I think the $24$ comes from perhaps his $z_k$ being of unit magnitude. $\endgroup$ – Ron Gordon Mar 19 '13 at 18:57

OK, I am convinced that your mistake lies in calling one of the poles $z=e^{-i \pi/3}$. For your keyhole contour, it should be $z=e^{i 5 \pi/3}$.

To see this, I am just going to write down the equation you get after applying the contour and the residues:

$$-\int_0^{\infty} \frac{dx}{x^3+8} = \frac{\log{2}+i \pi/3}{12 e^{i 2 \pi/3}} + \frac{\log{2}+i \pi}{12} + \frac{\log{2}+i 5\pi/3}{12 e^{i 4 \pi/3}}$$

You can show that the $\log{2}$'s cancel. The rest of it is

$$\frac{i \pi}{12} \frac{1}{3} [ e^{-i 2 \pi/3} + 3 + 5 e^{i 2 \pi/3}] = \frac{i \pi}{12} \frac{1}{3} (-3+3+i 2 \sqrt{3}) = -\frac{\pi}{6 \sqrt{3}}$$

The result follows.

  • $\begingroup$ Yep, that's basically what I got once I worked it out. I'm not sure why I wrote $-\pi/3$ as the other pole, since I had been using $5\pi/3$ instead. But I appreciate the effort! $\endgroup$ – Emily Mar 19 '13 at 19:49
  • $\begingroup$ Oh, I know why -- when I wrote the question I copy-pasta'd from a prior result :) $\endgroup$ – Emily Mar 19 '13 at 19:50

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