# Showing that $\Gamma(x) \Gamma(y) = \Gamma(x+y)\int\limits_0^1 \lambda^{x-1}(1-\lambda)^{y-1}d\lambda$

On page 56 of Titchmarsh's Theory of Functions, Titchmarsh makes the following claim:

\begin{align} \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} &= \phi(x,y); (x>0,y>0) \end{align} where \begin{align} \phi(x,y) &=\int\limits_0^\infty\frac{v^{y-1}}{(1+v)^{x+y}}dv\\ &= 2 \int\limits_0^{\frac{1}{2}\pi} (\cos \theta)^{2x-1}(\sin\theta)^{2y-1}d\theta\\ &= \int\limits_0^1 \lambda^{x-1}(1-\lambda)^{y-1}d\lambda \end{align}

I understand how the first integral is achieved using substitution, but I can't figure out how to go from the first integral to the second or the second integral to the third.

• Search for the derivation of the relation between beta and gamma functions. – StubbornAtom Jul 16 '18 at 4:32
• Second to third is by the substitution $\lambda = (\cos x)^{2}$ – Kavi Rama Murthy Jul 16 '18 at 5:55
• – StubbornAtom Jul 16 '18 at 6:15

## 1 Answer

From the first to second,

\begin{align} \phi(x,y) &=\int\limits_0^\infty\frac{v^{y-1}}{(1+v)^{x+y}}dv\\ &=\int\limits_0^\infty\frac{(\tan^2\theta)^{y-1}}{(1+(\tan^2\theta))^{x+y}}d(\tan^2\theta)\\ &=\int\limits_0^{\frac{1}{2}\pi}\frac{(\tan^2\theta)^{y-1}}{(\sec^2\theta)^{x+y}}(2\tan\theta\sec^2\theta)d\theta\\ &=2\int\limits_0^{\frac{1}{2}\pi}(\sin\theta)^{2y-2+1}(\cos \theta)^{-(2y-2+1)+(2x+2y)-2}d\theta\\ &= 2 \int\limits_0^{\frac{1}{2}\pi} (\cos \theta)^{2x-1}(\sin\theta)^{2y-1}d\theta\\ \end{align}

Now can you think of the next substitution? (Hint: look at the limits and the exponents)