Cahen–Mellin integral

Remarks

For real $y > 0$ this is the inverse Mellin transform of $\Gamma = \mathcal{M}[e^{-x}]$, so it follows from the Mellin inversion theorem. For complex $y$, put $\abs{\arg y} = \theta < \pi/2$; then $\abs{y^{-c - it}} = \abs{y}^{-c} e^{t \arg y} \le \abs{y}^{-c} e^{\theta \abs{t}}$, while $\abs{\Gamma(c + it)}$ decays like $\abs{t}^{c - 1/2} e^{-\pi \abs{t}/2}$ by Stirling's formula. So the integral converges absolutely and locally uniformly, it is holomorphic in $y$, and the identity extends from $y > 0$ by the identity theorem. A direct proof moves the line of integration to the left: by the residue theorem, the poles of $\Gamma$ at $s = -k$ with residues $(-1)^k / k!$ give $\sum_{k \ge 0} (-1)^k y^k / k! = e^{-y}$. The hypothesis $c > 0$ is needed: for $-1 < c < 0$ the line passes to the left of the pole at $0$, and the integral equals $e^{-y} - 1$.

Going deeper

For a student

This integral is the prototype of Mellin–Barnes integrals and of the method of shifting contours. In analytic number theory it is applied to $y = n x$ and summed over $n$: the sum $\sum_n a(n) e^{-n x}$ then becomes $\frac{1}{2\pi i} \int \Gamma(s) F(s) x^{-s} \, ds$ with $F$ the Dirichlet series of the $a(n)$. Shifting the line past the poles of $\Gamma(s) F(s)$ gives asymptotic expansions of the exponential sum as $x \to 0^+$. It is the smooth counterpart of Perron's formula, which uses the kernel $x^{s}/s$ instead of $\Gamma(s) x^{-s}$ and has worse convergence.