# Functional analysis proof of Ramanujan's Master Theorem

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).

• I remember the first time I saw this kind of heuristic derivation, it was with Euler–Maclaurin formula. here too it's enlightening. I think Berndt gave a 'rigourous proof' of Ramanujan's master theorem. Commented Jan 1, 2017 at 8:33
• I don't think it is "obviously false as written". Because $f$ is uniquely determined by the sequence $(\psi (k))_k.$ Any sequence $(A_k)_k$ is equal to $(\psi (k))_k$ for some $\psi$ integrable from $0$ to $\infty.$ Some restrictions needed are that the radius of convergence of the series for $f$ is $\infty$, and that $x^{n-1}f(x)$ is integrable from $0$ to $\infty$ for every positive integer $n.$ Commented Jan 1, 2017 at 10:23
• @user254665 The point is that $\psi$ is not uniquely determined by $f$ Commented Jan 1, 2017 at 16:50
• I think it is uniquely determined by $f$. Commented Nov 4, 2018 at 10:53
• The formula of Ramanujan is based on Mellin transform and a proof was given by G H Hardy in Ramanujan : Twelve Lectures on Subjects Suggested by His Life and Work. See page $189$. Commented Nov 5, 2018 at 9:00

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 $$Re(z), with poles at the points $$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).

• Sir. I looked up your research gate but could not find the paper you mention. A reference to a rigorous derivation of Ramanujan’s theorem would be amazing! Commented Dec 4, 2021 at 21:32
• @FShrike. See at researchgate.net/publication/235945980_Thesis_NBagis Commented Dec 6, 2021 at 6:28
• Many thanks, I’ll have a go at reading this Commented Dec 6, 2021 at 6:49

The shift operator $$e^{mt}$$ directly multiplies the Laplace transformed coeficient function $$\hat{a}(t)$$. Compare equations (D.4) and (L.3) This shift property provides a proof. $$..$$ Given: \begin{align} I(m)=\int_{0}^{\infty} z^{m-1} \ f(z) \ dz \tag{G.1} \\ f(z)= \sum_{s=0}^{\infty} \ \frac {\phi (s) \ (-z)^{s}}{s!} \tag{G.2} \\ Domain \ \phi (s)= \{ s\in R : s\geq -m \} \tag{G.3} \\ e^{-u}= \sum_{s=0}^{\infty} \ \frac{(-u)^{s}} {s!} \tag{G.4} \\ \Gamma (m) = \int_{0}^{\infty} u^{m-1} \ e^{-u} \ du \tag{G.5} \end{align} Define: \begin{align} a(s)=\phi (s-m) \tag{D.1} \\ \hat{a}(t)= \int_{0}^{\infty} e^{-ts} a(s) \ ds \tag{D.2} \\ \hat{a}(\lambda + i \omega) = \int_{0}^{\infty} e^{- i \omega s} e^{- \lambda s} a(s) \ ds \tag{D.3} \\ a(s)= \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} e^{st} \ \hat{a}(t) \ dt \tag{D.4} \\ v=t; \ u= e^{t}*z \tag{D.5} \\ dv \wedge du = e^{t} \ dt \wedge dz \tag{D.6} \end{align} Assumptions: \begin{align} \exists \lambda \ for \ \int_{0}^{\infty} e^{- i \omega} e^{- \lambda s} a(s) \ ds < \infty \tag{A.1} \\ \exists \lambda \ for \ a(s)= \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} e^{st} \ \hat{a}(t) \ dt \tag{A.2} \end{align} Lemmas: \begin{align} a(s+m)= \phi(s) \tag{L.1} \\ a(0)= \phi(-m) \tag{L.2} \\ a(s+m)= \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} e^{mt} e^{st} \ \hat{a}(t) \ dt \tag{L.3} \\ a(0)= \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} \hat{a}(t) \ dt \tag{L.4} \end{align} Lemma proofs: L.1: rewrite using D.1; L.2: rewrite using D.1; L.3: rewrite using D.4; L.4: rewrite using D.4. $$..$$ Proof: \begin{align} I(m)= & \int_{0}^{\infty} z^{m-1} \ f(z) \ dz \tag{P.1} \\ & = \int_{0}^{\infty} z^{m-1} \ \sum_{s=0}^{\infty} \ \frac {\phi (s) \ (-z)^{s}}{s!} \ dz \tag{P.2} \\ & = \int_{0}^{\infty} z^{m-1} \ \sum_{s=0}^{\infty} \ \frac {a(s+m) \ (-z)^{s}}{s!} \ dz \tag{P.3} \\ & = \int_{0}^{\infty} \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} z^{m-1} \ \sum_{s=0}^{\infty} \ \frac {e^{mt} e^{st} \hat{a}(t) (-z)^{s}}{s!} \ dt \ dz \tag{P.4} \\ & = \int_{0}^{\infty} \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} \ \hat{a} (t) \sum_{s=0}^{\infty} \ \frac{ (e^{t}z)^{m-1} (- e^{t} z)^{s}}{s!} \ (e^{t} dt \ dz) \tag{P.5} \\ & = \int_{0}^{\infty} \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} \ \hat{a}(v) \sum_{s=0}^{\infty} \ \frac {(u)^{m-1} (- u)^{s}}{s!} \ dv \ du \tag{P.6} \\ & = \left( \frac{1}{2 \pi i} \int_{\lambda - i \infty}^{\lambda + i \infty} \ \hat{a}(v) \ dv \right) \left( \int_{0}^{\infty} \sum_{s=0}^{\infty} \ \frac{ (u)^{m-1} (- u)^{s}}{s!} \ du \right) \tag{P.7} \\ & = a(0) \left( \int_{0}^{\infty} \sum_{s=0}^{\infty} \ \frac{ (u)^{m-1} (- u)^{s}}{s!} \ du \right) \tag{P.8} \\ & = \phi(-m) \left( \int_{0}^{\infty} \sum_{s=0}^{\infty} \ \frac{ (u)^{m-1} (- u)^{s}}{s!} \ du \right) \tag{P.9} \\ & = \phi(-m) \left( \int_{0}^{\infty} u^{m-1} e^{-u} \ du \right) \tag{P.10} \\ & = \phi(-m)\Gamma(m) \tag{P.11} \end{align} Proof steps: P.2: rewrite using G.2; P.3: rewrite using L.1; P.4: rewrite using L.3; P.5: reorder the factors; P.6: rewrite using D.5 and D.6; P.7: reorder the factors; P.8: rewrite using L.4; P.9: rewrite using L.2; P.10: rewrite using G.4, P.11: rewrite using G.5. $$..$$ Comments: For equation (A.1), $$\lambda$$ must be large enough to make the integral convergent. For equation (A.2), $$\lambda$$ must be to the right of all singularities of $$\hat{a}(t)$$ in the complex plane. If we interpret equation (D.3) as a Fourier transform, compute the inverse Fourier transform, we obtain equation (D.4). For detailed proof and discussion of equation (D.4) see: Bromwich integral (Inverse Laplace Transform) page 696 in MATHEMATICAL METHODS IN THE PHYSICAL SCIENCES Third Edition MARY L. BOAS.