# Integration involving beta and hypergeometric functions/ differential binomial

So, hey, everybody!

I have to integrate this $$\int_0^2 \sqrt[3]{\frac{x^2}{2-x}} \, dx$$ and I've already figured out that due to Chebyshev's theorem it cannot be done in terms of elementary functions, since we can rewrite the task as $$\int_0^2 x^{\frac{2}{3}}\left(2-x\right)^{-\frac{1}{3}} \, dx$$ and $$\frac{2}{3}+1-\frac{1}{3}=\frac{4}{3}$$ which is, obviously, not an integer number.

But, the point is that I'm not quite familiar with Beta-function and stuff like that, so, how do you write the answer to this problem?

Mathematica show something like that $$\frac{\Gamma\left[\frac{1}{2} \right]\Gamma\left[\frac{2}{3} \right] }{\Gamma\left[\frac{7}{6} \right] }$$ The question is: "Hey, how did Mathematica do this?"

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Here's a way to solve by using beta and gamma functions.

Let $$I=\int_0^2 \frac{x^{2/3}}{(2-x)^{1/3}}\,dx$$ The above is equivalent to: $$I=\int_0^2 \frac{(2-x)^{2/3}}{x^{1/3}}\,dx$$ Add the two expressions for $I$ to obtain: $$2I=\int_0^2 \frac{2}{x^{1/3}(2-x)^{1/3}}\,dx \Rightarrow I=\int_0^2\frac{1}{x^{1/3}(2-x)^{1/3}}\,dx=2\int_0^1 \frac{dx}{x^{1/3}(2-x)^{1/3}}$$ $$\Rightarrow I=2\int_0^1 \frac{dx}{(1-x)^{1/3}(1+x)^{1/3}}=2\int_0^1 (1-x^2)^{-1/3}\,dx$$ Use the substitution $x^2=t \Rightarrow dx=\frac{1}{2}t^{-1/2}\,dt$ to get: $$I=\int_0^1 t^{-1/2}(1-t)^{-1/3}=B\left(\frac{1}{2},\frac{2}{3}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)}$$ You can even simplify it more by writing $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ using Euler's reflection formula.

In a similar way, you can prove the results presented by Claude Leibovici.

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This is a very elegant solution ! – Claude Leibovici Jun 4 '14 at 12:23
@ClaudeLeibovici: Thank you! :) – Pranav Arora Jun 4 '14 at 12:23
Thanks:з Exactly what I needed – Lebesgue Jun 4 '14 at 12:49
I was preparing a so similar answer that it is no longer usefull to post it ! This gives me the opportunity to greet Claude Leibovici. – JJacquelin Jun 4 '14 at 13:05
By the way, I should like to reassure Lebesgue about the use of the Beta function with which he says to be not quite familiar. There is nothing mysterious : Special functions are just like more usual functions and you can find definitions and properties in the mathematical handbooks. Only with a bit of learning and training, one can practice them as fluently as exponentiel, logarithm, sinusoidal and other elementary functions. A review for general public (pp.18-36) : fr.scribd.com/doc/14623310/… – JJacquelin Jun 4 '14 at 13:13

It is relative easy with the change of variable $x=2t$ and the formulas for the Beta function from http://en.wikipedia.org/wiki/Beta_function $$\int_0^2 x^{\frac{2}{3}}\left(2-x\right)^{-\frac{1}{3}} \, dx = \int_0^1 (2t)^{\frac{2}{3}}\left(2-2t\right)^{-\frac{1}{3}} 2\, dt = 2^{1+\frac{2}{3}-\frac{1}{3}} \int_0^1 t^{\frac{2}{3}}\left(1-t\right)^{- \frac{1}{3}} \, dt =2^{\frac{4}{3}} B\left(\frac{5}{3},\frac{2}{3}\right)$$ The last expression is the same as the Mathematica $\Gamma$ term.

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I am not sure that this is the answer you expect but CAS are able to compute the antiderivative and find $$\int \sqrt[3]{\frac{x^2}{2-x}} \, dx=\frac{3}{4} \sqrt[3]{\frac{x^2}{2-x}} \left(2^{2/3} \sqrt[3]{2-x} \, _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{5}{3};\frac{x}{2}\right)+x-2\right)$$ If we compute the integral $$I(a)=\int_0^a \sqrt[3]{\frac{x^2}{2-x}} \, dx$$ there are very few cases where the result is quite nice. The only ones I found are $$I(1)=\frac{3}{4} \left(\frac{\sqrt{\pi } \Gamma \left(\frac{5}{3}\right)}{\Gamma \left(\frac{7}{6}\right)}-1\right)$$ $$I(2)=\frac{\sqrt{\pi } \Gamma \left(\frac{2}{3}\right)}{\Gamma \left(\frac{7}{6}\right)}$$

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