$\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$ when $z_k$ solves $z^3+8=0$ 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?
 A: 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.
Consider
$$\oint_C \frac{dz}{z^3+8}$$
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.
A: 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.
