Definite integration involving log

Question

Please help me to solve following definite integral:

$$ \int_{0}^{1} {\sqrt[3]{x\log\frac{1}{x}}dx}$$

Is there a reason you think there is a nice closed form? Is this an exercise? — Thomas Andrews, Jul 21 '17 at 16:40
http://www.wolframalpha.com says: -(Gamma[-(2/3)]/(2 6^(2/3))) — Mariusz Iwaniuk, Jul 21 '17 at 16:42
Wolfram Alpha gives the value as $$-\frac{\Gamma\left(\frac{-2}{3}\right)}{2\cdot 6^{2/3}}\approx 0.608494$$ — Thomas Andrews, Jul 21 '17 at 16:54

score 2 · Answer 1 · answered Jul 21 '17 at 17:08

If $x=e^{-y}$ then $dx=-e^{-y}\,dy$ and $$\sqrt[3]{x\log\frac{1}{x}}=e^{-y/3}\sqrt[3]y$$

So this is $$\int_{0}^{\infty} y^{1/3}e^{-4y/3}\,dy$$

Letting $y=\frac{3}{4}w$ then this is: $$\begin{align}\left(\frac 3{4}\right)^{4/3}\int_{0}^{\infty}w^{1/3}e^{-w}\,dw&=\left(\frac 3{4}\right)^{4/3}\Gamma\left(\frac{4}{3}\right)\\ &=\frac{1}{3}\left(\frac 3{4}\right)^{4/3}\Gamma\left(\frac{1}{3}\right)\\ &=\frac{6^{1/3}\Gamma\left(\frac{1}{3}\right)}{8} \end{align}$$

You get the same answer as Wolfram alpha if you use $\Gamma\left(\frac13\right)=-\frac{2}{3}\Gamma\left(-\frac23\right).$

Definite integration involving log

1 Answers1