alternate integral representation of beta function (2)

Substitute $x:=\frac{1}{1+s}$, $dx=\frac{-1}{{(1+s)}^2}ds$:

${\displaystyle {\int }_0^1}{x}^{p-1}{(1-x)}^{q-1}𝑑x$$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{1}{{(1+s)}^{p+1}}}{\left({\displaystyle \frac{s}{1+s}}\right)}^{q-1}𝑑s$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{s^{q-1}}{{(1+s)}^{p+q}}}𝑑s$

Since $B(p,q)=B(q,p)$ this gives:

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{s^{p-1}}{{(1+s)}^{p+q}}}𝑑s$$={\displaystyle \frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{\mathrm{\Gamma }(p+q)}}$