Gamma function as Mellin transform of the exponential

Example

Let $f(x) = e^{-x}$ for $x > 0$. Its Mellin transform is the gamma function: $$\mathcal{M}f(s) = \int_0^\infty x^{s-1} e^{-x} \, dx = \Gamma(s), \qquad \operatorname{Re} s > 0.$$ The right side is Euler's integral, which is the usual definition of $\Gamma$ on the half-plane $\operatorname{Re} s > 0$.

Check of Mellin convergence: let $\sigma = \operatorname{Re} s$. On $(0, 1]$, $x^{\sigma - 1} e^{-x} \le x^{\sigma - 1}$, which is integrable exactly when $\sigma > 0$. On $[1, \infty)$, $x^{\sigma - 1} e^{-x} \le C_\sigma e^{-x/2}$ for a constant $C_\sigma$, which is integrable for every $\sigma$. For $\sigma \le 0$, $x^{\sigma - 1} e^{-x} \ge e^{-1} x^{\sigma - 1}$ on $(0, 1]$, which is not integrable. So $f$ is Mellin convergent exactly on the half-plane $\operatorname{Re} s > 0$.

Integration by parts gives $\Gamma(s + 1) = s \Gamma(s)$, and $\Gamma(1) = \int_0^\infty e^{-x} \, dx = 1$, so $\Gamma(n + 1) = n!$. By the dilation rule, $\int_0^\infty x^{s-1} e^{-ax} \, dx = a^{-s} \Gamma(s)$ for $a > 0$. By the Mellin inversion theorem, for every $c > 0$ and $x > 0$, $$e^{-x} = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \Gamma(s) x^{-s} \, ds.$$