Help me to finish calculating $\int_0^{\infty} \frac{1}{x^3-1}dx$

We can in fact evaluate the Cauchy principal value of the integral as follows (which is what I think you were trying to do).

Consider the following contour integral:

$$\oint_C dz \frac{\log{z}}{z^3-1}$$

$C$ is a modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. The modification lies on small semicircular bumps above and below $z=1$ of radius $\epsilon$, and we will consider the limits as $\epsilon \to 0$ and $R\to\infty$.

Let's evaluate this integral over the contours. There are $8$ pieces to evaluate, as follows:

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log{x}}{x^3-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1+\epsilon}^R dx \frac{\log{x}}{x^3-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log{\left (R e^{i \theta}\right )}}{R^3 e^{i 3 \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}\right ) }+i 2 \pi}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (\epsilon e^{i \phi}\right )}}{\epsilon^3 e^{i 3 \phi}-1} $$

(To see this, draw the contour out, including the bumps about $z=1$.)

As $R \to \infty$, the fourth integral vanishes as $\log{R}/R^2$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^2)$, while the eighth integral vanishes as $\epsilon \log{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become

$$PV \int_0^{\infty} dx \frac{\log{x} - (\log{x}+i 2 \pi)}{x^3-1} + \frac{2 \pi^2}{3}$$

It should be appreciated that, in the fifth, sixth, and seventh integrals, the $i 2 \pi $ factor appears because, on the lower branch of the real axis, we write $z=x \, e^{i 2 \pi}$. In the sixth integral, in fact, $z = e^{i 2 \pi} + \epsilon \, e^{i \phi + 2 \pi}$.

The $PV$ denotes the Cauchy principal value of the integral. As it stands, the integral does not actually converge. Nevertheless, we are not actually considering the integral straight through the pole at $z=1$, but a very small detour around the pole. Thus, in the limit, we get the Cauchy PV. A little rearranging cancels the $\log$ term, and we now have:

$$-i 2 \pi PV \int_0^{\infty} \frac{dx}{x^3-1} + \frac{2 \pi^2}{3}$$

The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles. The poles here are at $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$. Note that the pole at $z=1$ is not inside the contour $C$ because of the detour around that "pole". It should be appreciated that the poles must have their arguments between $[0,2 \pi]$ because of the way we defined $C$.

In any case, we now have that the above 1D integrals over the positive real line are equal to

$$i 2 \pi \left [\frac{i 2 \pi/3}{3 e^{i 4 \pi/3}} + \frac{i 4 \pi/3}{3 e^{i 8 \pi/3}} \right ] = \frac{2 \pi ^2}{3}+i \frac{2\pi ^2}{3 \sqrt{3}} $$

We may now solve for the principal value and get:

$$ PV \int_0^{\infty} \frac{dx}{x^3-1} = -\frac{\pi}{3 \sqrt{3}} $$

Using partial fractions directly I think it is simpler:

$$\int\frac1{x^3-1}dx=\frac13\left(\int\frac1{x-1}-\frac{x+2}{x^2+x+1}\right)dx=$$

$$\frac13\left(\log(x-1)-\frac12\int\frac{2x+1}{x^2+x+1}dx-\frac32\int\frac1{\left(x+\frac12\right)^2+\frac34}dx\right)=$$

$$=\frac13\log(x-1)-\frac12\log(x^2+x+1)-\sqrt3\int\frac{\frac2{\sqrt3}dx}{1+\left(\frac{2x+1}{\sqrt3}\right)^2}=$$

$$=\log\frac{\sqrt[3]{x-1}}{\sqrt{x^2+x+1}}-\arctan\frac{2x+1}{\sqrt3}+C , $$

and etc.

• Subdivide $(0,\infty)$ into $(0,1)$ and $(1,\infty)$.
• On the latter, let $t=\dfrac1x$.
• $a^3-b^3=\big(a-b\big)\Big(a^2+ab+b^2\Big)$.
• Complete the square in the denominator.