# definite integral $\int_{0}^{1}\sqrt[3]{x\log\frac 1x}$ [duplicate]

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

## marked as duplicate by Guy Fsone, projectilemotion, Jack, Error 404, NosratiOct 23 '17 at 13:36

• $\log$ is $\ln$ or $\log_{10}$? – Holo Oct 23 '17 at 11:40
• – Guy Fsone Oct 23 '17 at 11:43
• It must be a coincidence: same two users answer the same question twice. – Jack Oct 23 '17 at 13:02
• It is while answering that I remember that I have answered typical question in the pass then I went through my answers list . See that I am the first to flag this post as duplicate – Guy Fsone Oct 23 '17 at 13:50

With substitutions $\ln\dfrac1x=u$ and then $\dfrac43u=v$ we have $$I=\int_0^\infty u^\frac13e^{-\frac43u}du=\left(\dfrac34\right)^\frac43\int_0^\infty v^\frac13e^{-v}dv=\left(\dfrac34\right)^\frac43\Gamma\left(\dfrac43\right)$$
• $\Gamma$ is the upper gamma function yes? – Holo Oct 23 '17 at 11:39
set $u = -\ln x$ then $dx -e^{-u} du$
$$I=\int_0^1 \sqrt[3]{x\log{\frac{1}x}} \, \mathrm dx$$
then set $v= 4/3u$ so that $$I=3/4\int_0^\infty e^{-v} (3/4v)^{1/3}dv = {\left(\frac{3}{4}\right)}^{4/3}\int_0^\infty e^{-v} v^{1/3}dv ={\left(\frac{3}{4}\right)}^{4/3}\Gamma(\frac{4}{3}) = {\left(\frac{3}{4}\right)}^{4/3}\frac{1}{3}\Gamma(\frac{1}{3}).$$