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.

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

1 Answer 1


This can be easily done using beta function.

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
    $\begingroup$ oh, I understand !! One more Ques, how do you come up with the idea I(m,n) ? thank u $\endgroup$
    – Jaytone
    Sep 18, 2021 at 10:48
  • 2
    $\begingroup$ @Jaytone I have encountered a lot of such integrals. So, experience. BTW, if you are satisfied with the answer, consider accepting it. $\endgroup$ Sep 18, 2021 at 10:49
  • 1
    $\begingroup$ (+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 $B(-\mu,\mu+1)$. $\endgroup$ Sep 18, 2021 at 11:01
  • $\begingroup$ @projectile that's also a good way to do it. $\endgroup$ Sep 18, 2021 at 11:14
  • $\begingroup$ Very nice. I was not expecting such a general result. $\endgroup$
    – K.defaoite
    Sep 18, 2021 at 11:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .