# 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...

$$\frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}=\frac{2}{5}B(1/2,1/6)=\frac{2}{5} \int_0^{\infty} \frac{ \mathrm dt}{\sqrt{t}(1+t)^{2/3}}$$

...?

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Is your question, why does the first integral equal the second? or is your question, why does the beta function equal the second integral? or is your question, why does the beta function equal that expression in pi and Gamma? In short, what parts do you know, and what parts not? –  Gerry Myerson Jun 17 '12 at 1:26

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?

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Around here might be there. –  Did Jun 17 '12 at 6:20
@did Yep, that's it! –  Pedro Tamaroff Jun 17 '12 at 12:24
The expression of the beta function as an integral from $0$ to $\infty$ is exactly what I am trying to use to show the result, but I don't know how to link it to $\int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx$ –  Chon Jun 17 '12 at 12:42
@Chon Try for example, $x+1 =\frac{\tau +1}{\tau}$ or any similar rational transformation that will keep the limits in $0,1$ or $0, \infty$, maybe. –  Pedro Tamaroff Jun 17 '12 at 17:46

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?

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Hmm, can you explain how you went from second to third step? I am somewhat confused as to how you pulled that $y$ out so cleanly. –  TenaliRaman Jun 17 '12 at 9:32
@TenaliRaman:Well,it is just substitution of variables:let $x=\sqrt{y} u$,then $\int_{0}^{\infty}(x^2+y)^{-2/3}dx=\int_{0}^{\infty}(u^2+1)^{-2/3}y^{-2/3}\sqrt{‌​y}du$. –  y zhao Jun 17 '12 at 10:35
Ah, that didn't strike me at all, I thought you did some clever algebraic trick to get y out :-). Thanks for the explanation (+1)! –  TenaliRaman Jun 17 '12 at 10:56