## Proof of the Week for 20120716

Prove that Euler‘s original form of the gamma function, $\Gamma\left(z\right) = \int_0^1\left[\ln\left(\frac{1}{t}\right)\right]^{z-1}dt$ is equivalent to the more common definition: $\Gamma\left(z\right) = \int_0^\infty t^{z-1}e^{-t}dt$.

Let us begin by rewriting Euler’s form as $\Gamma\left(z\right) = \int_0^1\left[-\ln t\right]^{z-1}dt$. We then integrate by substitution by letting $u=-\ln t$, which can be expressed as $t = e^{-u}$. After adjusting the limits of integration, we have $\Gamma\left(z\right) = \int_\infty^0 u^{z-1}\left(-e^{-u}\right)du$ which becomes $\Gamma\left(z\right) = \int^\infty_0 u^{z-1}e^{-u}du$