Proof of beta function using convolution

Prove the following

$$\beta(x, y)=\int^{1}_{0}t^{x-1}\, (1-t)^{y-1}\,dt= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma{(x+y)}}$$

$$\textit{proof}$$

Let us choose some functions $f$ and $g$

$$f(t) = t^{x} \,\, , \, g(t) = t^y$$

Hence we get

$$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds $$

So by the convolution rule we have the following

$$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y ) $$

We can now use the Laplace of the power

$$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$

Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$

$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{-1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!}
{(x+y+1)!}$$

So we have the following

$$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$

By definition we have

$$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds
$$

Now put \( t=1 \) we get

$$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$

By using that \( n! = \Gamma{(n+1)}\)  we deduce that

$$ \int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma
{(x+y+2)}}$$

which can be written as

$$ \int^{1}_0 s^{x-1}(1-s)^{y-1}\,ds= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma
{(x+y)}}$$

This entry was posted in Beta function, Gamma function and tagged , , , . Bookmark the permalink.

Leave a Reply