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

Question

The problem is apparently simple: Find $$\displaystyle{\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x }$$

However, I have not been able to get around it. The usual "big circle" strategy is useless since we don't have parity. With substitution I was able to prove that, choosing $t = 1/x$, we have $$ \int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_1^\infty \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t $$ And, analogously $$ \int_1^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_0^1 \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t $$ However, this doesn't seem to be helpful. I could now write my integral as \begin{align} \int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x= & \int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x + \int_1^\infty \!\! \frac{\log x}{1+ x^3} \,\mathrm{d} x \\ = &\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x - \int_0^1 \!\!\frac{x \log x}{1+ x^3} \,\mathrm{d} x \\ = &\int_0^1 \!\!\frac{1-x}{1+ x^3} \log x \,\mathrm{d} x\\ \end{align} This is a proper definite integral and I hoped it could be useful, e.g. to use integration by series But I couldn’t manage it.

Answer 1

We shall use the residue theorem to compute. Consider the keyhole contour,

and put $f(z)=\frac{\log^2(z)}{1+z^3}$. It is easy to estimate that the integral on the big circle and the small circle vanishes when $R\to\infty$ and $r\to 0$. Compute the residue at $z=e^{\frac{\pi i}{3}},e^{\pi i},e^{\frac{5 \pi i}{3}}$ and the sum of these residues is $\frac{4}{27} \left(1-3 i \sqrt{3}\right) \pi ^2$. According to the former discussion,
$$2\pi i\cdot \frac{4}{27} \left(1-3 i \sqrt{3}\right) \pi ^2=-4\pi i\int^\infty_0\frac{\log x}{1+x^3}\mathrm{d}x+4\pi^2\int^\infty_0\frac{\mathrm{d}{x}}{1+x^3}.$$ Hence $$ \int^\infty_0\frac{\log x}{1+x^3}\mathrm{d}x=-\frac{2 \pi ^2}{27}$$

Answer 2

Recalling the evaluation of the Beta function $\int_0^1 t^{a-1} (1-t)^{b-1} \mathrm{d}t = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, we see that for $0 < \alpha < \beta$ and $\frac{\alpha}{\beta} \notin \mathbb{Z}$ we can evaluate the following integral $$ \int_0^{\infty} \frac{x^{\alpha -1}}{1+x^{\beta}}\,\mathrm{d}x \overset{1-t = \frac{1}{1 + x^{\beta}}}{=} \frac{1}{\beta}\int_0^{1} t^{\frac{\alpha}{\beta}-1}(1-t)^{-\frac{\alpha}{\beta}}\,\mathrm{d}x = \frac{1}{\beta}\Gamma\left(\frac{\alpha}{\beta}\right)\Gamma\left(1-\frac{\alpha}{\beta}\right)= \frac{\pi}{\beta} \csc\left( \frac{\pi \alpha}{\beta}\right) $$ where on the last step we used Euler's reflection formula $\Gamma(1-z)\Gamma(z)= \pi \csc(\pi z)$. Differentiating the above equation w.r.t. $\alpha$ on both sides gives $$ \int_0^{\infty} \frac{\ln(x)x^{\alpha -1}}{1+x^{\beta}}\,\mathrm{d}x = - \frac{\pi^2}{\beta^2} \csc\left( \frac{\pi \alpha}{\beta}\right) \cot\left( \frac{\pi \alpha}{\beta}\right) \tag{1} $$ And substituting $\alpha = 1$ and $\beta =3$ gives your desired integral $$ \int_0^{\infty} \frac{\ln(x)}{1+x^{3}}\,\mathrm{d}x = - \frac{\pi^2}{9} \csc\left( \frac{\pi }{3}\right) \cot\left( \frac{\pi }{3}\right) = \boxed{- \frac{2\pi^2}{27}} $$

Answer 3

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

\begin{align} \int_0^\infty&\frac{\ln x}{1+x^3}{d}x = \int_0^1\frac{(1-x)\ln x}{1+x^3}{d}x \\ &=\int_0^1\frac{\ln x}{1+x} - \underset{x^3\to x}{\frac{x^2\ln x}{1+x^3}}\ {dx} = \frac89\int_0^1\frac{\ln x}{1+x}{d}x =-\frac{2\pi^2}{27} \end{align}

Answer 4

If you do not want tou use residues, you have to play with the roots of unity.

Making the problem more general, compute the antiderivative $$I=\int \frac{\log (x)}{1+x^n}\,dx=\int \frac{\log (x)}{\prod _{i=1}^n (x-r_i)}\,dx=\sum_{i=1}^n a_i \int\frac{\log (x)}{x-r_i}\,dx$$ $$\int\frac{\log (x)}{x-r_i}\,dx=\text{Li}_2\left(\frac{x}{r_i}\right)+\log (x) \log \left(1-\frac{x}{r_i}\right)$$

Recombine everything and use the bounds.

In fact, there is a nice formula for any $n>2$. Try to find it.

Answer 5

$$I=\int_{0}^{\infty} \frac{\ln x}{1+x^3} dx$$ Let $x=e^t$, then $$I=\int_{-\infty}^{\infty} \frac{t e^t dt}{1+e^{3t}}=\int_{-\infty}^{0} \frac{t e^t dt}{1+e^{3t}}+\int_{0}^{\infty} \frac{t e^t dt}{1+e^{3t}}=-\int_{0}^{\infty} \frac{t e^{-t} dt}{1+e^{-3t}}+\int_{0}^{\infty} \frac{t e^{-2t} dt}{1+e^{-3t}}$$ Using IGP expansion, we can write $$I=\int_{0}^{\infty} \sum_{0}^{\infty} (-1)^n[-te^{-(3n+1)t}+te^{(3m+2)t} ] dt =\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$$ Using $\psi^1(z)=\sum_{n=0}^{\infty}\frac{1}{(n+z)^2}$ $$\implies I=\frac{1}{36}\left([\psi^1(1/3)-\psi^{1}(5/6)]-[\psi^1(1/6)-\psi^1(2/3)]\right)$$ Using $\psi^1(z)+\psi^1(1-z)=-\pi^2\csc^2(\pi z)$ see https://en.wikipedia.org/wiki/Polygamma_function , we get $$I=\frac{1}{36}[\frac{4\pi^2}{3}-4\pi^2]=-\frac{2\pi^2}{27}.$$