TITLE: Functional analysis proof of Ramanujan's Master Theorem
QUESTION [19 upvotes]: According to mathworld, Ramanujan's master theorem is the statement that if $$f(z) = \sum_{k=0}^{\infty} \frac{\phi(k) (-z)^k}{k!}$$ for some function (analytic or integrable) $\phi$, then $$\int_0^{\infty} x^{n-1} f(x) \, \mathrm{d}x = \Gamma(n) \phi(-n).$$
As written it is obviously false as the values of an (analytic or integrable) function $\phi$ at natural numbers do not determine its values anywhere else. However it turns out that $$\int_0^{\infty} x^{s-1} f(x) \, \mathrm{d}x = \Gamma(s) \phi(-s)$$ for arbitrary $s$ under growth conditions on $\phi$.
Recently I came across an elementary "proof": if $T$ denotes the shift operator $T\phi(s) := \phi(s+1),$ then we can write $$f(z) = \sum_{k=0}^{\infty} \frac{(-z)^kT^k \phi(0)}{k!} = e^{-zT}\phi (0)$$ such that $$\int_0^{\infty} x^{n-1} f(x) \, \mathrm{d}x = \int_0^{\infty} x^{n-1} e^{-xT} \phi(0) \, \mathrm{d}x = \Gamma(n) T^{-n}\phi(0) = \Gamma(n) \phi(-n),$$ by plugging $T$ into the Gamma integral $$\int_0^{\infty} x^{n-1} e^{-xs} \, \mathrm{d}x = \Gamma(n) s^{-n}.$$ I am curious whether this argument can be made rigorous with functional analysis on an appropriate function space (which necessarily would have to have some growth conditions).

REPLY [8 votes]: Write
$$
(M\Psi)(s)=\int^{\infty}_{0}\Psi(t)t^{s-1}dt,
$$
where $s\in A_{\Psi}$ with
$$
A_{\Psi}:=\left\{s\in\textbf{C}:\int^{\infty}_{0}\Psi(t)t^{s-1}dt<\infty\right\}.
$$
Then we have the next Mellin inversion formula
$$
\Psi(z)=\frac{1}{2\pi i}\int^{\sigma+i\infty}_{\sigma-i\infty}(M\Psi)(s)z^{-s}ds,
$$
where $\sigma\in Re\left(A_{\Psi}\right)$.
Theorem 1. If $\Psi(z)$ have power series arround $0$ with radious of convergence $r>0$ and if $x\in\textbf{R}$ such that
$$
\int^{\infty}_{0}\left|\Psi(t)\right|t^{x-1}dt<+\infty.
$$
Then the Mellin transform of $\Psi$ can be analyticaly continued to a meromorphic function in the halphplane $Re(z)<x$, with poles at the points $z=-m$, $m$ is non-negative integers such $m>-x$.
Theorem 2. Let $x>0$ and $f,\Psi$ analytic in $\textbf{C}$ and satisfining the condition
$$
|f(z)(M\Psi)(x+iz)|\leq C (1+|z|)^{\lambda}e^{-\delta |Re(z)|},\tag 1
$$
for all $z$ such that $Im(z)\geq 0$ and $c,\lambda,\delta>0$ constants, with the condition that $|z|=x+N+1/2$, where $N-$natural number saficiently large. Then the integral
$$
\int^{\infty}_{-\infty}f(t)M\Psi(x+it)dt
$$
converges absolutely, the series
$$
\sum^{\infty}_{m=0}\frac{\Psi^{(m)}(0)}{m!}f(i(x+m))
$$
converges in Abel sence and
$$
\int^{\infty}_{-\infty}f(t)(M\Psi)(x+it)dt=2\pi\lim_{r\rightarrow 1}\sum^{\infty}_{m=0}\frac{\Psi^{(m)}(0)}{m!}f(i(x+m))r^m.\tag 2
$$
Moreover if
$$
\left|\frac{\Psi^{(m)}(0)}{m!}f(i(x+m))\right|\leq \frac{C'}{m+1},
$$
then (2) converges.
As application of the above theorem we have
Theorem 3. Let $f$ be analytic in the upper half plane $Im(z)>0$ and continuous in $Im(z)\geq 0$ and such that
$$
|f(z)|\leq C (1+|z|)^{\rho}\left(\frac{|z|}{e}\right)^{Im(z)}e^{b|Re(z)|},
$$
in $Im(z)\geq 0$, where $0\leq b\leq \pi/2$. Then for $x>0$ we have
$$
\int^{\infty}_{-\infty}f(t)\Gamma(x+it)dt=2\pi\lim_{r\rightarrow1}\sum^{\infty}_{m=0}\frac{(-1)^m}{m!}f(i(x+m))r^m
$$
Theorem 3 is Ramanujan's Master Theorem.
I have full proofs of that in my PhD thesis but there are written in Greek (see Researchgate: Nikos Bagis).