# Beta in terms of power of t over power of (1+t)

The following formula holds: $$B(x,y)=\displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}} \mathrm{d}t,$$ where $B$ denotes the beta function.
From the definition, $$B(x,y)=\displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u.$$ We will proceed using substitution. Let $u=\dfrac{t}{t+1}$ so that $\mathrm{d}u=\dfrac{1}{(t+1)^2} \mathrm{d}t$. Since $u=0$ means $t=0$ and $u=1$ means $t=\infty$, we get $$\begin{array}{ll} B(x,y) &= \displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u \\ &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( 1 -\dfrac{t}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( \dfrac{1}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ &= \displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}}, \end{array}$$ as was to be shown.