How to evaluate integral using complex analysis How can I use complex analysis to find the following integral:
$$ \int_{0}^{\infty} \frac{\log x \ dx }{{(1+x^3) }^2} $$
Can you suggest a proper contour and hints to tackle further integration.
 A: The contour you can take is this

Hints for the calculations
Your integral is of the form
$$\int_0^{+\infty} R(x)\log(x)\ \text{d}x$$
We choose the contour integration $\Gamma(R, \epsilon)$ as showed above, and we notice that
$$\lim_{x\to \infty} xR(x) = 0$$
$R(x)$ has no poles for $x\geq 0$ so we can proceed. Also, our path is chosen with $0 < \theta < 2\pi$ for $\arg(z)$.
$$\log(z) = \log|z| + i\theta ~~~~~~~~~~~~~ \theta = \arg(z)$$
To solve it, we consider the integral of $R(x)\log^2(x)$ instead.
Along that path, $\log^2(z) $ has no singularity inside $\Gamma(R, \epsilon)$ and also
$$zR(z) \log^2(z) \to 0 ~~~~~~~ |z| \to \infty$$
Because the degree of $R(z)$ is at least greater than $2$ "points" with respect to the numerator.
Also
$$zR(z)\log^2(z) \to 0 ~~~~~~~ z\to 0$$
So we have:
$$2\pi i\ \sum \ \text{Res}\ (R(z)\log^2(z)) = \left(\int_{L^+} + \int_{L^-}\right) R(z)\log^2(z)\ \text{d}z$$
On $L^+$ we have $z = x e^{i0^+}$ hence
$$\log z = \log z + i0^+ = \log x$$
and on $L^-$ we have
$$\log z = \log x + i2\pi$$
So
$$\left(\int_{L^+} + \int_{L^-}\right) R(z)\log^2(z)\ \text{d}z = \int_0^{+\infty} R(x)\left[\log^2 x - (\log x - 2\pi i)^2\right]\ \text{d}x$$
$$ = 4\pi^2\int_0^{+\infty} R(x)\ \text{d}x - 4\pi i\int_0^{+\infty} R(x)\log(x)\ \text{d}x$$
Hence we end up with the important formula
$$\int_0^{+\infty} R(x)\log(x)\ \text{d}x = -\frac{1}{2}\Re\left[\sum \ \text{Res}\ (R(z)\log^2(z))\right]$$
So what you need to do is just to compute the residues of
$$\frac{\log^2(z)}{(1+z^3)^2}$$
Hints for the residues
Notice that
$$1+z^3 = (z+1)(z^2-z+1)$$
Poles are
$$z_0 = e^{\pi i} = -1$$
$$z_1 = e^{i\pi/3}$$
$$z_2 = e^{5i\pi/3}$$
Of you may prefer to calculate the roots of $z^2-z+1$, it's the same.
Notice that all your residues are of order two, because thou have
$$(1+z^3)^2 \longrightarrow ((z+1)(z^2-z+1))^2 = (z-1)^2(z^2-z+1)^2$$
Evaluate them, follow the rule above, take the real part and you will get the result
$$\boxed{-\frac{4\pi^2}{81} - \frac{2\sqrt{3}\pi}{27}}$$
A: Hint:  Choose the contour as:
$C=[-R,-r]\cup[r, R]\cup C_R\cup \gamma_r$, where $C_R$ is the upper half circle with radius of $R$ and $\gamma_r$ is the small upper half circle with radius of $r<1$ bypassing $0$.
Prove that on $C_R$ and $\gamma_r$ integral approaches $0$ as $R\to \infty$ and $r\to0$. Note that on $[-R,-r], \: \log(-x)=\log x+\pi i$.
