Integral representation of the Beta function

Question

How do I get from $ \int_{0}^{1}{t^{x-1} (1-t)^{y-1}\,\mathrm dt} $ to $ \displaystyle\int_{0}^{\infty}{ \frac{t^{x-1} + t^{y-1}}{(1+t)^{x+y}} \,\mathrm dt} $ ? I have tried different change of variables but without success.

Answer 1

Substitute

$$\frac1u=t\implies dt=-\frac{du}{u^2}\implies$$

$$\int_0^1 t^{x-1}(1-t)^{y-1}dt=\int_\infty^1u^{1-x}\left(1-\frac1u\right)^{y-1}\left(-\frac{du}{u^2}\right)=\int_1^\infty u^{1-x}u^{1-y}(u-1)^{y-1}u^{-2}du=$$

$$=\int_1^\infty u^{-(x+y)}(u-1)^{y-1}du=:I$$

and now substitute

$$u=z+1\implies du=dz\implies I=\int_0^\infty(z+1)^{-(x+y)}z^{y-1}dz=$$

$$=\frac12\int_0^\infty\frac{z^{x-1}+z^{y-1}}{(z+1)^{x+y}}dz$$