Applications of Ramanujan's Master Theorem

Question

Ramanujan's Master Theorem is really neat. Unfortunately, however I have only used it once before, and I want to use it more. I would like a list of integrals to which I may apply this beautiful theorem.

The Theorem: (Taken from Wikipedia)

If $f(x)$ is a complex valued function with a series representation in the form $$f(x)=\sum_{n\geq0}\frac{\phi(n)}{n!}(-x)^n$$ Then $$\int_0^\infty x^{s-1}f(x)\mathrm dx=\Gamma(s)\phi(-s)$$ Where $\Gamma(s)$ is the Gamma function.

Cheers!

Answer 1

As promised within the comment section a little collection of integrals I either solved myself using the RMT or encountered while searching for some

\begin{align*} &(1)&&\int_0^\infty x^{s-1}\sin(x)\mathrm dx=\Gamma(s)\sin\left(\frac{\pi s}2\right)\\ &(2)&&\int_0^\infty x^{s-1}\cos(x)\mathrm dx=\Gamma(s)\cos\left(\frac{\pi s}2\right)\\ &(3)&&\int_0^\infty \frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac\pi{\sin(\pi \alpha)}\\ &(4)&&\int_0^\infty x^{s-1}\log(1+x)\mathrm dx=\frac1s\frac\pi{\sin(\pi s)}\\ &(5)&&\int_0^\infty x^{s-1}~_2F_1(\alpha,\beta;\gamma;-x)\mathrm dx=B(a,s-\alpha)\frac{\Gamma(\beta)\Gamma(s-\beta)}{\Gamma(s-\gamma)\Gamma(\gamma)}\\ &(6)&&\int_0^\infty \frac{\operatorname{Li}_3(-x)}{1+x}x^{s-1}\mathrm dx=\frac\pi{\sin(\pi s)}[\zeta(3)-\zeta(3,1-s)]\\ &(7)&&\int_0^\infty \log^m(x)\sin(x^n)\mathrm dx~=~\lim_{\phi\to0}\frac{\mathrm d^m}{\mathrm d\phi^m}\left[\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right)\right]\\ &(8)&&\int_0^\infty \sin(x^n)\mathrm dx=\sin\left(\frac\pi{2n}\right)\Gamma\left(1+\frac1n\right)\\ &(9)&&\int_0^\infty \cos(x^n)\mathrm dx=\cos\left(\frac\pi{2n}\right)\Gamma\left(1+\frac1n\right)\\ &(10)&&\int_0^\infty \frac{\mathrm dx}{1+x^n}=\frac\pi n\csc\left(\frac\pi n\right) \end{align*}

Feel free to ask for clarification if some of them are not clear at all. I will see whether I can find some more $($also I want to refer to the collection of integrals within this article again$)$.

Answer 2

The modified Mellin transform (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,

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

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

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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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{1}{(1-x)^{n}} \;\frac{t^n}{n!} = e^{\frac{1}{1-x}t} , $

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

$\; \; \; \; \;\tilde{f}(s) = (1-x)^{s}$ for $ Re(s) > 0.$

I include this last example because it is important in characterizing the distributional expansion of the Heaviside step function and its derivatives--the derivatives of the Dirac delta function--in terms of the superposition of the Laguerre polynomials as demonstrated in the MO-Q "What's the matrix of the logarithm of the derivative operator, $\ln(D)$? What is the role of this operator in various math fields?"

Some other related MO-Qs:

"An analytic continuation of power series coefficients"

"Deriving the functional equation for ζ(s) from summing the powers of the zeros required to count the integers"

"Newton series and Fourier transform - is there an analogy?"