Gammazzz. Topic: Calculus/S&S.

Definition: (Gamma function) $ \displaystyle \Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t} dt = \lim_{n \rightarrow \infty} \frac{n! \cdot n^z}{z(z+1)(z+2) \cdots (z+n)} $. Note that the Gamma function satisfies the property $ \Gamma(z+1) = z\Gamma(z) $ and $ \Gamma(k+1) = k! $ for nonnegative integers $ k $.

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Problem: Prove that $ \Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin{\pi z}} $.

Solution: We use the second definition for $ \Gamma $ given above to evaluate

$ \displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{n! \cdot n^z}{z(z+1)(z+2) \cdots (z+n)} \cdot \lim_{n \rightarrow \infty} \frac{n! \cdot n^{-z}}{-z(-z+1)(-z+2) \cdots (-z+n)} $.

Divide the top and bottom by $ n! $ in both limits to obtain

$ \displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{ n^z}{z(1+\frac{z}{1})(1+\frac{z}{2}) \cdots (1+\frac{z}{n})} \cdot \lim_{n \rightarrow \infty} \frac{n^{-z}}{-z(1-\frac{z}{1})(1-\frac{z}{2}) \cdots (1-\frac{z}{n})} $.

Multiplying these limits together, we get

$ \displaystyle \Gamma(-z) \Gamma(z) = \lim_{n \rightarrow \infty} \frac{ 1}{-z^2(1+\frac{z}{1})(1-\frac{z}{1})(1+\frac{z}{2})(1-\frac{z}{2}) \cdots (1+\frac{z}{n})(1-\frac{z}{n})} $.

Notice, however, that

$ \displaystyle \sin{\pi z} = \pi z \cdot \prod_{n=1}^{\infty} \left(1+\frac{z}{n}\right)\left(1-\frac{z}{n}\right) $

because it has zeros at all of the integers and has first term $ \pi z $ (by Taylor series). So

$ \displaystyle \Gamma(-z) \Gamma(z) = \frac{ 1}{-z \cdot \frac{\sin{\pi z}}{\pi}} $

and

$ \displaystyle -z \Gamma(-z) \Gamma(z) = \frac{\pi}{\sin{\pi z}} $.

Lastly, by the property $ \Gamma(1-z) = -z\Gamma(-z) $, we arrive at the identity

$ \Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin{\pi z}} $.

QED.

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Comment: The Gamma function is quite a useful function and actually shows up a lot in higher-level math. The proof above assumes the equivalence of the two forms of the function, which can be proven using the fact that the Gamma function is log-convex. In fact, in can be shown that only one such function exists that satisfies $ f(x+1) = xf(x) $ and is log-convex.

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Practice Problem: Show that $ \Gamma(x) $ is log-convex for $ x > 0 $, i.e. $ \log{\Gamma(tx_1+(1-t)x_2)} \le t\log{\Gamma(x_1)}+(1-t)\log{\Gamma(x_2)} $ for all $ 0 \le t \le 1 $.