Evaluate integral $\int_0^\infty \frac{\log x}{1+x^3}dx$

Question

I can't seem to evaluate this integral

$$\int_0^\infty \frac{\log (x)}{1+x^3}\, dx$$

No matter what I do, I tried every method I can think of but nothing works, I tried substituting $x$ with mutiple thing, integration by parts... I tried using the app called WolframAlpha and it says that my integral evaluates to $-\frac{2\pi^2}{27}$, but it doesn't show the step-by-step process. Some help would be appreciated.

Answer 1

An elementary solution

\begin{align}\int_0^\infty\frac{\ln x}{x^3+1}{d}x = & \int_0^1\frac{(1-x)\ln x}{x^3+1}{d}x =\int_0^1\frac{\ln x}{x+1}{d}x - \int_0^1{\frac{x^2\ln x}{x^3+1}} \overset{x^3\to x}{dx}\\ =&\> \left(1-\frac19\right)\int_0^1\frac{\ln x}{x+1}{d}x =\frac89\left(-\frac{\pi^2}{12}\right) =-\frac{2\pi^2}{27} \end{align}

where $\int_0^1\frac{\ln x}{x+1}{d}x=-\frac{\pi^2}{12}$

Answer 2

We shall just use an elementary well-known result for which you can find several proofs involving the beta function or Ramanujan’s master theorem:

$$\int_{0}^{\infty} \frac{x^{s-1}}{x^n+1} \, dx=\frac{\pi}{n} \csc\left(\frac{\pi s}{n}\right)$$

Take $n=3$ and the partial derivative with respect to $s$:

$$\int_{0}^{\infty}\frac{x^{s-1} \ln x}{x^3+1}\, dx=-\frac{\pi^2}{9} \cot \left(\frac{\pi s}{3}\right)\csc\left(\frac{\pi s}{3}\right)$$

Taking $s\to 1$ gives the required result:

$$\boxed{\int_{0}^{\infty} \frac{\ln x}{x^3+1} \, dx=-\frac{2\pi^2}{27}}$$