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

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