Gamma functions are useful in proving the following result.
Exponential mean and variance
If \(X\) has an \(\ExponDistn(\lambda) \) distribution with pdf
\[ f(x) \;\;=\;\; \lambda e^{-\lambda x} \]then its mean and variance are
\[ E[X] \;\;=\;\; \frac 1 {\lambda} \spaced{and} \Var(X) \;\;=\;\; \frac 1 {\lambda^2} \](Proved in full version)