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.

  • $\begingroup$ This page is what you are looking for en.wikipedia.org/wiki/… $\endgroup$
    – b00n heT
    Nov 23, 2016 at 14:36
  • 1
    $\begingroup$ the approach i used here will work:math.stackexchange.com/questions/1859034/… $\endgroup$
    – tired
    Nov 23, 2016 at 14:40
  • $\begingroup$ by the way i would recommend a beta function apporach instead of contour integration. seems to be less cumbersome $\endgroup$
    – tired
    Nov 23, 2016 at 15:01

2 Answers 2


The contour you can take is this

enter image description here

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.


$$zR(z)\log^2(z) \to 0 ~~~~~~~ z\to 0$$

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

$$\log z = \log x + i2\pi$$


$$\left(\int_{L^+} + \int_{L^-}\right) R(z)\log^2(z)\ \text{d}z = \int_0^{+\infty} R(x)\left[\log^2 x - (\log x - 2\pi i)^2\right]\ \text{d}x$$

$$ = 4\pi^2\int_0^{+\infty} R(x)\ \text{d}x - 4\pi i\int_0^{+\infty} R(x)\log(x)\ \text{d}x$$

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


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

$$\boxed{-\frac{4\pi^2}{81} - \frac{2\sqrt{3}\pi}{27}}$$

  • 1
    $\begingroup$ Thanks a ton! Even though I nearly reached this, I was stuck up finding the value of $4\pi^2\int_0^{+\infty} R(x)\ \text{d}x$.I am trying again to figure out how it turns to be 0. $\endgroup$ Nov 24, 2016 at 14:23
  • 1
    $\begingroup$ Oh! I got it. As $4\pi^2\int_0^{+\infty} R(x)\ \text{d}x$ when divided by $4\pi i $ gives me a imaginary number, hence doesn't contribute to the result. That solves the problem! $\endgroup$ Nov 24, 2016 at 14:38

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 $[-R,-r], \: \log(-x)=\log x+\pi i$.


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