show $B(x,y+1) = \frac{y}{x+y}B(x,y)$ with integration by parts

Question

Hi I have to show that $B(x,y+1) = \frac{y}{x+y}B(x,y)$ by doing integration by parts on $B(x,y+1) = \int_{0}^{1} t^{x+y-1} (\frac{1-t}{t}) ^ {y} dt$. But however I try I always have this integral that doesnt vanish. Is there any trick here? Any hint would be enough, thanks!

Answer 1

We have $$B(x,y)=\int_{0}^{1}t^{x+y-2}\big(\frac{1}{t}-1\big)^{y-1}dt$$

Integrating by parts we have $$B(x,y+1) = \int_{0}^{1} t^{x+y-1} (\frac{1-t}{t}) ^ {y} dt$$ $$=\big[\big(\frac{1}{t}-1\big)^{y}\frac{t^{x+y}}{x+y}\big]_{t=0}^{t=1}+\frac{y}{x+y}\int_{0}^{1}t^{x+y-2}\big(\frac{1}{t}-1\big)^{y-1}dt$$

and the first term vanishes. Can you end it?

Answer 2

By partial integration we get: $B(x,y+1)=\int_0^1 t^{x+y-1}(1-\frac{1}{t})^y dt=\int_0^1 (1-\frac{1}{t})^y d\frac{t^{x+y}}{x+y}=(1-\frac{1}{t})^y\frac{t^{x+y}}{x+y}|_0^1-\int_0^1\frac{t^{x+y}}{x+y}y(1-\frac{1}{t})^{y-1}\frac{1}{t^2}dt=\frac{y}{x+y}\int_0^1t^{x+y}(t-1)^{y-1}\frac{1}{t^{y+1}}dt=\frac{y}{x+y}\int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)$