Computing $ \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx$ 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...
$$\frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}=\frac{2}{5}B(1/2,1/6)=\frac{2}{5} \int_0^{\infty} \frac{ \mathrm dt}{\sqrt{t}(1+t)^{2/3}}$$
...?
 A: 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?
A: I think that follows from a clever transformation of the Beta function I saw some days ago around here:
Let 
$$B(x,y)=\int_0^1 t^{x-1} (1-t)^{y-1} dt$$
Set $t = \dfrac{1}{\tau +1}$, so that
$$\eqalign{
  & B(x,y) =  - \int_\infty ^0 {{{\left( {\frac{1}{{\tau  + 1}}} \right)}^{x - 1}}} {\left( {1 - \frac{1}{{\tau  + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau  + 1} \right)}^2}}}  \cr 
  & B(x,y) = \int_0^\infty  {{{\left( {\frac{1}{{\tau  + 1}}} \right)}^{x - 1}}} {\left( {\frac{\tau }{{\tau  + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau  + 1} \right)}^2}}} = \int_0^\infty  {\frac{{{\tau ^{y - 1}}}}{{{{\left( {\tau  + 1} \right)}^{x + y}}}}d\tau }  \cr} $$
Similarily, let
$$\eqalign{
  & t = \frac{\tau }{{\tau  + 1}}  \cr 
  & dt = \frac{{d\tau }}{{{{\left( {\tau  + 1} \right)}^2}}} \cr} $$
$$\eqalign{
  & B(x,y) = \int_0^\infty  {{{\left( {\frac{\tau }{{\tau  + 1}}} \right)}^{x - 1}}} {\left( {1 - \frac{\tau }{{\tau  + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau  + 1} \right)}^2}}}  \cr 
  & B(x,y) = \int_0^\infty  {{{\left( {\frac{\tau }{{\tau  + 1}}} \right)}^{x - 1}}} {\left( {\frac{1}{{\tau  + 1}}} \right)^{y - 1}}\frac{{d\tau }}{{{{\left( {\tau  + 1} \right)}^2}}} = \int_0^\infty  {\frac{{{\tau ^{x - 1}}}}{{{{\left( {\tau  + 1} \right)}^{x + y}}}}d\tau }  \cr} $$
Can you use that to show the result?
