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I have to calculate the following integral using beta and gamma functions: $$ \int\limits_0^1 \frac{x\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}} $$

I came up with this terrible solution. Firstly, let's break it into two parts: $$ \int\limits_0^1 \frac{(x-2)\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}} + \int\limits_0^1 \frac{2\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}} $$

The first one is $-B\left(\frac 13,\frac 23\right)$. The second one can be simplified with substitution $x = 1 - \frac 1t$ to $$ 2\int\limits_{-\infty}^0 \frac{dt}{(t+1)(t-1)^\frac 23} $$

But it's too unwieldy in my opinion. Furthermore, it's not so easy to evaluate $B\left(\frac 13,\frac 23\right)$. Is there any easier solution?

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  • $\begingroup$ Is this an exercise? $\endgroup$ – science Feb 7 '15 at 22:46
  • $\begingroup$ Yes, it's an exercise. $\endgroup$ – perlik Feb 7 '15 at 23:11
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    $\begingroup$ In general, for $a\in(0,1)$ we have $~\displaystyle\int_0^1\frac{dx}{(1-ax)\sqrt[3]{x^2(1-x)}} ~=~ \frac{2\pi}{\sqrt3~\sqrt[3]{1-a}}.~$ In this case, $a=\dfrac12.$ $\endgroup$ – Lucian Jan 1 '16 at 8:13
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For the second integral, make the substitution $ \displaystyle x = \frac{u}{1+u}$.

Then

$$ \begin{align} 2 \int_{0}^{1} \frac{dx}{(2-x) \sqrt[3]{x^2(1-x)}} &= 2 \int_{0}^{\infty} \frac{du}{(2+u)u^{2/3}} \\ &= \int_{0}^{\infty} \frac{du}{\left(1+ \frac{u}{2}\right)u^{2/3}} \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{1}{(1+t)t^{2/3}} \, dt \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{t^{1/3-1}}{(1+t)^{1/3+2/3}} \, dt \\ &= \frac{2}{2^{2/3}} B \left(\frac{1}{3}, \frac{2}{3} \right) \tag{1} \\&= \frac{2}{2^{2/3}} \frac{\Gamma \left(\frac{1}{3} \right) \Gamma \left(1- \frac{1}{3} \right)}{\Gamma(1)} \\ &= \frac{2}{2^{2/3}} \frac{2 \pi}{\sqrt{3}} \tag{2} \\ &= \frac{2^{4/3} \pi}{\sqrt{3}}. \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Beta_function#Properties

$(2)$ https://en.wikipedia.org/wiki/Gamma_function#General

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