We define for all real $x > 0$ the sequence $(J_{n}(x))_{n \geqslant 0}$ by: $$J_{n}(x) = \int_{0}^{1} (1-t)^{n} t^{x-1} dt$$ Show that, for all integers $n$, $n \geqslant 0$, $$\forall x > 0, \quad J_{n+1}(x) = \frac{n+1}{x} J_{n}(x+1)$$
We define for all real $x > 0$ the sequence $(J_{n}(x))_{n \geqslant 0}$ by:
$$J_{n}(x) = \int_{0}^{1} (1-t)^{n} t^{x-1} dt$$
Show that, for all integers $n$, $n \geqslant 0$,
$$\forall x > 0, \quad J_{n+1}(x) = \frac{n+1}{x} J_{n}(x+1)$$