Show that $T(x+1)=xT(x)$ if $T(x)=\int _0 ^\infty e^{-t}t^{x-1} dt$

Question

I need to calculate the following: the integral is from 0 to infinity

I am given that $ T(x) = \int \limits _0 ^\infty e^{-t}t^{x-1} dt$.

I need to show that $ T(x+1)=xT(x) $.

So my logic was to plug $x+1$ for $x$ which gives

$ T(x)=\int \limits _0 ^\infty e^{-t} t^x dt $

However I am not quite sure how to integrate this, I have an idea that it should be integration by parts but I am not quite sure where to go from here.

Answer 1

Of course you need $x > 0$ for the improper integral $\int_0^\infty e^{-t} t^{x-1}\; dt$ to converge. This is the definition of the Gamma function.

Yes, you can use integration by parts. In the integral for $T(x+1)$, try $u = t^x$ and $dv = e^{-t}\; dt$.