Ramanujan's Master Formula: A proof and relation to umbral calculus

Question

The Ramanujan's master theorem states that: $$ \int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}a_nx^ndx=\Gamma(s)a_{-s} $$

I found a really strange proof recently on a personal blog: Define $$ \tau *a_n=a_{n+1} $$ we have $$ {\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}a_nx^n\text{dx} \\=a_0\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1*\tau_a*x)^n}{n!}\text{dx} \\=a_0\int_0^{\infty}x^{s-1}\text{e}^{-\tau_ax}\text{dx} \\=\frac{a_0}{\tau_a^s}\int_0^{\infty}{(\tau_a x)}^{s-1}\text{e}^{-\tau_ax}\mathrm{d\tau_a x} =\frac{a_0}{\tau_a}\Gamma(s) \\=a_{-s}\Gamma(s) } $$

My question is, how can I understand the above proof? Can anyone give a rigorous explanation for the above proof? For example, I totally have no idea about the meaning of $\color{blue}{\mathrm{d\tau_ax}}$(in the sense of analysis).

Would it be related to Umbral calculus?

Also, I want some further readings if possible.

Answer 1

The RMF is definitely related to umbral calculus via the modified Mellin transform (MMT) pair and symbolic extension of the iconic Euler gamma function integral. The proof you copied? I don't know. The MMT pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation.

First consider the MMT and its inverse

$$\tilde{f}(s) = MMT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$

$$f(x) = MMT^{-1}[\tilde{f}(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \tilde{f}(s) \frac{x^{-s}}{(-s)!} \; ds .$$

Then the RMF holds for a class of functions such that

$$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} = \sum_{n=0} \tilde{f}(-n) \frac{(-x)^n}{n!} \; ,$$

that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and other ranges of $x$.

Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation

$$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x)$$

reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation

$$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$

with $H(x)$ the Heaviside step function.

So, under the conditions above,

$$\tilde{f}(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$

and this suggests the analytic continuation and relation to umbral calculus

$$\tilde{f}(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$

The iconic guiding example is the Euler gamma function integral rep with $(a.)^n = a_n = c^n$

$$ (a.)^{-s} = a_{-s} = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}}.$$

Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials,

$$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$

$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z)$$

where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is

$$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$

Note that

$$B_n(z) = -n \; \zeta(1-n,z),$$

$$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) = (-1)^n B_n(0) = (-1)^n b_n.$$

Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) = s \; B_{s-1}(z)$.

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Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago.

For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences."

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Added 2/5/21:

Other examples of interpolation of $a_n$ for the exponential generating funcrtion $g(t) = e^{a.t}$ from the MMT of $f(t) = g(-t) = e^{-a.t}$, or, conversely, surmising the MMT of $f(t)$ from the Taylor series coefficients of $g(t)$ via $a_n \; |_{n \rightarrow -s} =a_{-s} =\tilde{f}(s)$:

1) $\;g(t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!} ),$

$\; \; \; \; \;\tilde{f}(s) =\cos(\pi \frac{s}{2})$ for $0 < Re(s) < 1,$

2) $\;g(t) = \sin(t)= \sum_{n \geq 0} \sin(\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \sin(-t) = \sum_{n \geq 0} \sin(-\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =-\sin(\pi \frac{s}{2})$ for $-1 < Re(s) < 1,$

3) $\;g(t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =(-s)! $ for $0 < Re(s) < 1,$

4) $\;g(t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s)=\cos(\pi s) (-s)!$ for $0 < Re(s) < 1,$

5) $\;g(t) = \ln(1-t) = \sum_{n \geq 0} \; -(n-1)! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = \ln(1+t) = -\sum_{n \geq 0} \cos(\pi n) \; (n-1)! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = -(-s-1)! $ for $-1 < Re(s) < 0,$

6) $\;g(t) =\sum_{n \ge 0} \frac{x^n}{n!} \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = J_0(2 \sqrt{xt}) =\sum_{n \ge 0} (-1)^n \frac{x^n}{n!} \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = \frac{x^{-s}}{(-s)!}$ for $0 < Re(s) < \frac{3}{4},$

7) $\;g(t) = e^{-t^2} =\sum_{n \ge 0} \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!} \; \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = e^{-t^2},$

$\; \; \; \; \;\tilde{f}(s) = \cos(\pi\frac{ s}{2}) \; \frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} \;$ for $ Re(s) > 0.$

8) $\;g(t) =\sum_{n \ge 0} \; \frac{a_{\bar{n}}\; b_{\bar{n}}}{c_{\bar{n}}} \; \frac{t^n}{n!} = F(a,b;c;t)$, the hypergeometric function, where, e.g.,

$\; \; \; \; \;a_{\bar{n}} = \frac{(a+n-1)!}{(a-1)!}$, the rising factorial,

$\; \; \; \; \;f(t) = g(-t) = F(a,b;c;-t),$

$\; \; \; \; \;\tilde{f}(s) = \frac{a_{-\bar{s}} \; b_{-\bar{s}}}{c_{-\bar{s}}},$ see the Mellin-Barnes contour integral.

Answer 2

The rigorous proof of Ramanujan master theorem is the easiest part, the hard one is to find the precise conditions we need, ie. in which cases it starts to fail. I'll rename $\Gamma(s)\phi(-s)$ into $F(s)$.

If $F(s)$ is analytic and has exponential decay on a strip $\Re(s)\in (a,b),b >a>0$ then (Cauchy integral formula) so does $F''(s)$ on $\Re(s)\in (a+\epsilon,b-\epsilon)$, the inverse Mellin transform $f(x)=M^{-1}[F(s)](x)=\frac1{2i\pi}\int_{c-i\infty}^{c+i\infty} F(s)x^{-s}ds, c\in (a,b)$ converges and is analytic for $arg(x)\in (-r,r)$ and $x^{v-1}f(x)$ is integrable on $(0,\infty)$ for $v\in (a,b)$ so we have the convergent Mellin transform $F(s)=M[f(x)](s)=\int_0^\infty x^{s-1}f(x)dx$ for $\Re(s)\in (a,b)$. If also $f$ is equal to a power series on $(0,\delta)$ then $F(s)$ extends meromorphically to $\Re(s)<b$ with simple poles at $-n$ of residue $f^{(n)}(0)/n!$.

Forgetting about $F$, if we ask that $f(x)=\sum_{n\ge 0} \frac{a_n}{n!} x^n$ converges for $x\in(0,\delta)$ and extends analytically to $(0,\infty)$ such that $x^{b-1}f(x)$ is integrable for some $b>0$ then $F(s)=\int_0^\infty x^{s-1}f(x)dx$ is analytic for $\Re(s)\in (0,b)$ and it extends meromorphically to $\Re(s)<b$ with simple poles at negative integers of residue $a_n$.

On the space of such sequences $(a_n)$, with $(T^k a)_n= a_{k+n}$, the linear map $U(x)(a) = (e^{xT} a)(0)=\sum_{n\ge 0} \frac{a_n}{n!} x^n$ for $x\in (0,\delta)$ extends analytically to $(0,\infty)$ and the theorem is that the obtained linear map $V(s)(a)=\frac1{\Gamma(s)}\int_0^\infty x^{s-1}U(x)(a)dx,\Re(s)\in (0,b)$ extends analytically to $\Re(s)< b$ and $V(-k)=T^k$ for all integer $k\ge 0$.

Answer 3

This line of argument goes back to 1874 papers by J. W. L. Glaisher and J. O’Kinealy. A discussion of this early work and a critical examination (conditions on $a_n$ for which the theorem holds) is given in Ramanujan's Master Theorem, by Amdeberhan et al.

See also this related MSE discussion.