6
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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...

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

...?

  • 1
    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?2012-06-17
  • 0
    A [related question](http://math.stackexchange.com/questions/844216).2014-10-22

2 Answers 2