I am stuck with the following integral $\displaystyle{\int_0^1 \dfrac{x^a}{\sqrt{1-x}} dx}$. The only restriction I have is $a > -\dfrac{1}{2}$.
According to WolframAlpha, the answer is $\sqrt{\pi} \dfrac{\Gamma{(a+1)}}{\Gamma{\left(a+\dfrac{3}{2}\right)}}$ for Re ${a} > -1$. I know that $\sqrt{\pi} = \Gamma \left(\dfrac{1}{2} \right)$, but this does not really help with tackling the problem.
I also found the following expression in Gradshteyn&Ryzhik, Table of Integrals:
$\displaystyle{\int_0^{1} \dfrac{x^{n}}{\sqrt{1-x}} dx = 2 \dfrac{2n!!}{(2n+1)!!}}$, where !! is the double factorial.
I have tried integration by parts (unsuccessfully).
Any help, tricks or tips would be greatly appreciated.
Best Answer
$$ \int_0^1 \frac{x^a}{\sqrt{1-x}}dx=\int_0^1x^{a}(1-x)^{-\frac{1}{2}}dx=\beta\left(a+1,\frac{1}{2}\right) $$ where $\beta$ is the $\beta$ function. $\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ for all $x,y>0$ (see https://en.wikipedia.org/wiki/Beta_function#Relationship_to_the_gamma_function for a proof) therefore, using the fact that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$, we have $$ \int_0^1 \frac{x^a}{\sqrt{1-x}}dx=\sqrt{\pi}\frac{\Gamma(a+1)}{\Gamma\left(a+\frac{3}{2}\right)} $$