Verify that $Γ(x)$ = $(x − 1)Γ(x − 1)$ for all $x > 1$.

Question

$Γ(x)$ = $\int_0^{∞} e^{-t}t^{x-1}dt$

Plugging $(x-1)$ into this equation, I get

$Γ(x-1)$ = $\int_0^{∞} e^{-t}t^{x-2}dt$

Integrating by parts, I eventually end up with $-e^{-t}t^{x-1}]_0^∞$ + $(x-1)$$\int_0^{∞} e^{-t}t^{x-2}dt$

I can turn the right part into $(x-1)Γ(x-1)$, but that still leaves me with $-e^{-t}t^{x-1}]_0^∞$ + $(x-1)Γ(x-1)$. How do I get rid of the left side?

Answer 1

You have, for $x>1$, $$ \lim_{t\to \infty}e^{-t}t^{x-1}=0 $$ and $$ \lim_{t\to 0^+}e^{-t}t^{x-1}=0 $$ giving

$$ \left[-e^{-t}t^{x-1}\right]_0^∞=0. $$