# Calculate $\int_{0}^{\infty}{\frac{\ln{x}}{x^{\frac{2}{3}}(1+x)}}\mathrm dx$

I am having some difficulty with this exercise:

Calculate $$\int_{0}^{\infty}{\frac{\ln{x}}{x^{\frac{2}{3}}(1+x)}}\mathrm dx$$

By Wolfram Alpha, I know that the Answer is $$\frac{-2}{3}\pi^2$$. I tried to change variable $$\arctan(\sqrt{x})=t$$ cause I see (1+x), and it lead to $$\int_{0}^{\pi/2}{\frac{\ln({\tan(t)})}{\tan(t)^{1/3}}}\mathrm dx$$, and got stuck.

Can anyone help me solve this or give me some ideas.

• Partial fractions then integrate by parts? Sep 18, 2021 at 10:44

Starting with the substitution $$x\mapsto x^3$$, we get $$I = 9\int_0^\infty\frac{\ln x}{1+x^3}\,\mathrm dx$$ Now, consider the integral $$I (m,n)= \int_0^\infty \frac{x^m}{1+x^n}\,\mathrm dx$$ Using $$x^n=t$$, \begin{align} I(m,n) &= \frac1n\int_0^\infty\frac{t^{\frac{m+1}n-1}}{1+t}\,\mathrm dt \\ &= \frac1n \mathcal B\Big(\frac{m+1}n,1-\frac{m+1}n\Big)\\ &= \frac\pi n\csc \Big(\frac{\pi(m+1)}n\Big)\end{align} Differentiating w.r.t. $$m$$, $$\int_0^\infty\frac{x^m\ln x}{1+x^n}\,\mathrm dx = -\frac{\pi^2}{n^2}\csc \Big(\frac{\pi(m+1)}n\Big)\cot \Big(\frac{\pi(m+1)}n\Big)$$ Now, setting $$(m,n)=(0,3)$$ and multiplying by $$9$$, you will get the desired result.
• (+1) Nice answer! Alternatively, you can avoid the substitution $x\mapsto x^3$ by noting that $$\int_0^{\infty} \frac{\ln{x}}{x^{2/3}(1+x)}~dx=\int_{0}^{\infty}\frac{x^{-2/3}\ln{x}}{1+x}dx=\frac{\partial}{\partial \mu} \int_0^{\infty} \frac{x^{\mu}}{1+x}~dx\Big|_{\mu=-2/3}.$$ and the last integral is just $B(-\mu,\mu+1)$. Sep 18, 2021 at 11:01