I have to show the followiing equation, but I can't see how that can even be true:
$$\frac{B(x,y)}{c^{y}} = \int_{0}^{\infty} \frac{t^{x-1}}{{(c+t)}^{x+y}}$$
What I tried is
$$\int_{0}^{\infty} \frac{t^{x-1}}{{(1+t)}^{x+y}c^{y}} = \int_{0}^{\infty} \frac{t^{x-1}}{{(1+t)}^{x}{(1+t)}^{y}c^{y}} = \int_{0}^{\infty} \frac{t^{x-1}}{{(1+t)}^{x}{(c+ct)}^{y}}$$ but this doesn't seem to get me anywhere since it doesn't result in the integral on the right side.
We have $$c^{-y}B(x,y)=\int_{0}^{\infty}(c+t)^{-x-y}t^{x-1}dt \space\space(\dagger)$$
Differentiating wrt to $c$ we obtain (when we have uniform convergence) $$-yc^{-y-1}B(x,y)=\int_{0}^{\infty}(-x-y)(c+t)^{-x-y-1}t^{x-1}dt$$ $$=-(x+y)\int_{0}^{\infty}\frac{t^{x-1}}{(c+t)^{x+(y+1)}}\space\space(\star)$$
But we know from $(\dagger)$ that $$c^{-(y+1)}B(x,y+1)=\int_{0}^{\infty}\frac{t^{x-1}}{(c+t)^{x+(y+1)}}dt$$
So substituting this in $(\star)$ gives $$-yc^{-y-1}B(x,y)=\int_{0}^{\infty}(-x-y)(c+t)^{-x-y-1}t^{x-1}dt$$ $$=-(x+y)\int_{0}^{\infty}\frac{t^{x-1}}{(c+t)^{x+y+1}}dt=-(x+y)c^{-y-1}B(x,y+1)$$
Thus $$B(x,y)=\frac{x+y}{y}B(x,y+1)$$
