Ramanujan's master theorem

Remarks

This is the form of the theorem proved by G. H. Hardy. The idea is to write $f$ as a Mellin–Barnes integral: by the residue theorem, $f(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \frac{\pi}{\sin(\pi s)} \varphi(-s) x^{-s} \, ds$ for $0 < c < \delta$, since $\pi / \sin(\pi s)$ has residue $(-1)^k$ at $s = -k$. The growth bound with $A < \pi$ beats the decay $e^{-\pi \abs{w}}$ of $1/\sin(\pi s)$ and makes the integral converge; Mellin inversion then gives the formula. Writing $\varphi(k) = \lambda(k) / \Gamma(k + 1)$ and using $\Gamma(s)\Gamma(1 - s) = \pi / \sin(\pi s)$ gives the classical form: if $f(x) = \sum_{k \ge 0} \lambda(k) (-x)^k / k!$, then $\int_0^\infty x^{s-1} f(x) \, dx = \Gamma(s) \lambda(-s)$, under the corresponding conditions on $\lambda$. Example: $\varphi = 1$ gives $f(x) = 1/(1 + x)$ and $\int_0^\infty x^{s-1}/(1 + x) \, dx = \pi / \sin(\pi s)$ for $0 < \operatorname{Re} s < 1$. Example: $\lambda = 1$ gives $f(x) = e^{-x}$ and the formula $\int_0^\infty x^{s-1} e^{-x} \, dx = \Gamma(s)$. The growth condition is needed: $\varphi(z) = \sin(\pi z)$ vanishes at all integers, so $f = 0$, but $\varphi(-s) \ne 0$ for non-integer $s$; here $A = \pi$ and the bound with $A < \pi$ fails.

Going deeper

For a student

The growth conditions make the passage from integers to all complex numbers legitimate. By Carlson's theorem, a function of this type is determined by its values at the non-negative integers. The proof is an instance of the general Mellin method. A Mellin–Barnes integral, shifted to the left, gives the power series at $0$, and its inverse transform is the right side. The theorem yields at once the Mellin transforms of the binomial series $(1 + x)^{-a}$, of Bessel functions and of many hypergeometric functions.