TITLE: Computing $ \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx$
QUESTION [6 upvotes]: I would like to show that
$$ \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx = \frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}$$
thanks to the beta function which I am not used to handling...
$$\frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}=\frac{2}{5}B(1/2,1/6)=\frac{2}{5} \int_0^{\infty} \frac{ \mathrm dt}{\sqrt{t}(1+t)^{2/3}}$$
...?

REPLY [6 votes]: We can calculate the first integral by double integral.
1.Denote your first integral by $I$.Then
$\eqalign{
& I=2\int_{0}^{\infty}(x^2+1)^{1/3}-x^{2/3}dx \cr
&  =\frac{2}{3}\int_{0}^{1}\int_{0}^{\infty}(x^2+y)^{-2/3}dxdy \cr
&  =\frac{2}{3}\int_{0}^{1}y^{-1/6}dy\int_{0}^{\infty}\frac{1}{(x^2+1)^{2/3}}dx \cr
&  =\frac{4}{5}\int_{0}^{\infty}\frac{1}{(x^2+1)^{2/3}}dx}$
2.Use formula found by Peter Tamaroff.What would you see?