Gamma Function derivation using differentiation of definite integral?

Question

The usual derivation involves integration by parts technique, while I want a derivation using differentiation of definite integral. Is there a way? $$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt\,= (x-1)!$$ Wikipedia has the usual derivation using integration by parts, but can't this be done using differentiation of definite integral?

Answer 1

First note that $$\int_0^\infty e^{-tr}dt\,= \frac 1r$$

Now differentiate $n-1$ times with respect to $r$ (within the integral sign on the left) to obtain:$$(-1)^{n-1}\int_0^\infty t^{n-1}e^{-rt}dt\,= (-1)^{n-1} \frac {(n-1)!}{r^n}$$ and evaluate at $r=1$.