# Ramanujan's master theorem

In mathematics, Ramanujan's master theorem (named after Srinivasa Ramanujan) is a technique that provides an analytic expression for the Mellin transform of an analytic function.

The result is stated as follows:

If a complex-valued function $f(x)$ has an expansion of the form

$f(x)=\sum _{k=0}^{\infty }{\frac {\,\varphi (k)\,}{k!}}(-x)^{k}$ then the Mellin transform of $f(x)$ is given by

$\int _{0}^{\infty }x^{s-1}f(x)dx=\Gamma (s)\,\varphi (-s)$ where $\Gamma (s)$ is the gamma function.

It was widely used by Ramanujan to calculate definite integrals and infinite series.

Higher-dimensional versions of this theorem also appear in quantum physics (through Feynman diagrams).

A similar result was also obtained by Glaisher.

## Alternative formalism

An alternative formulation of Ramanujan's master theorem is as follows:

$\int _{0}^{\infty }x^{s-1}\left(\,\lambda (0)-x\,\lambda (1)+x^{2}\,\lambda (2)-\,\cdots \,\right)dx={\frac {\pi }{\,\sin(\pi s)\,}}\,\lambda (-s)$ which gets converted to the above form after substituting $\lambda (n)\equiv {\frac {\varphi (n)}{\,\Gamma (1+n)\,}}$ and using the functional equation for the gamma function.

The integral above is convergent for $0<{\mathcal {Re}}(s)<1$ subject to growth conditions on $\varphi$ .

## Proof

A proof subject to "natural" assumptions (though not the weakest necessary conditions) to Ramanujan's Master theorem was provided by G. H. Hardy employing the residue theorem and the well-known Mellin inversion theorem.

## Application to Bernoulli polynomials

The generating function of the Bernoulli polynomials $B_{k}(x)$ is given by:

${\frac {z\,e^{x\,z}}{\,e^{z}-1\,}}=\sum _{k=0}^{\infty }B_{k}(x)\,{\frac {z^{k}}{k!}}$ These polynomials are given in terms of the Hurwitz zeta function:

$\zeta (s,a)=\sum _{n=0}^{\infty }{\frac {1}{\,(n+a)^{s}\,}}$ by $~\zeta (1-n,a)=-{\frac {B_{n}(a)}{n}}~$ for $~n\geq 1~$ . Using the Ramanujan master theorem and the generating function of Bernoulli polynomials one has the following integral representation:

$\int _{0}^{\infty }x^{s-1}\left({\frac {e^{-ax}}{\,1-e^{-x}\,}}-{\frac {1}{x}}\right)dx=\Gamma (s)\,\zeta (s,a)\!$ which is valid for $~0<{\mathcal {Re}}(s)<1~$ .

## Application to the Gamma function

Weierstrass's definition of the Gamma function

$\Gamma (x)={\frac {\,e^{-\gamma \,x\,}}{x}}\,\prod _{n=1}^{\infty }\left(\,1+{\frac {x}{n}}\,\right)^{-1}e^{x/n}\!$ is equivalent to expression

$\log \Gamma (1+x)=-\gamma \,x+\sum _{k=2}^{\infty }{\frac {\,\zeta (k)\,}{k}}\,(-x)^{k}$ where $\zeta (k)$ is the Riemann zeta function.

Then applying Ramanujan master theorem we have:

$\int _{0}^{\infty }x^{s-1}{\frac {\,\gamma \,x+\log \Gamma (1+x)\,}{x^{2}}}\operatorname {d} x={\frac {\pi }{\sin(\pi s)}}{\frac {\zeta (2-s)}{2-s}}\!$ valid for $0<{\mathcal {Re}}(s)<1$ .

Special cases of $s={\frac {1}{2}}$ and $s={\frac {3}{4}}$ are

$\int _{0}^{\infty }{\frac {\,\gamma x+\log \Gamma (1+x)\,}{x^{5/2}}}\,\operatorname {d} x={\frac {2\pi }{3}}\,\zeta \left({\frac {3}{2}}\right)$ $\int _{0}^{\infty }{\frac {\,\gamma \,x+\log \Gamma (1+x)\,}{x^{9/4}}}\,\operatorname {d} x={\sqrt {2}}{\frac {4\pi }{5}}\zeta \left({\frac {5}{4}}\right)$ ## Application to Bessel functions

The Bessel function of the first kind has the power series

$J_{\nu }(z)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\Gamma (k+\nu +1)k!}}{\bigg (}{\frac {z}{2}}{\bigg )}^{2k+\nu }$ By Ramanujan's master theorem, together with some identities for the gamma function and rearranging, we can evaluate the integral

${\frac {2^{\nu -2s}\pi }{\sin {(\pi (s-\nu ))}}}\int _{0}^{\infty }z^{s-1-\nu /2}J_{\nu }({\sqrt {z}})\,dz=\Gamma (s)\Gamma (s-\nu )$ valid for $0<2{\mathcal {Re}}(s)<{\mathcal {Re}}(\nu )+{\tfrac {3}{2}}$ .

Equivalently, if the spherical Bessel function $j_{\nu }(z)$ is preferred, the formula becomes

${\frac {2^{\nu -2s}{\sqrt {\pi }}(1-2s+2\nu )}{\cos {(\pi (s-\nu ))}}}\int _{0}^{\infty }z^{s-1-\nu /2}j_{\nu }({\sqrt {z}})\,dz=\Gamma (s)\Gamma {\bigg (}{\frac {1}{2}}+s-\nu {\bigg )}$ valid for $0<2{\mathcal {Re}}(s)<{\mathcal {Re}}(\nu )+2$ .

The solution is remarkable in that it is able to interpolate across the major identities for the gamma function. In particular, the choice of $J_{0}({\sqrt {z}})$ gives the square of the gamma function, $j_{0}({\sqrt {z}})$ gives the duplication formula, $z^{-1/2}J_{1}({\sqrt {z}})$ gives the reflection formula, and fixing to the evaluable $s={\tfrac {1}{2}}$ or $s=1$ gives the gamma function by itself, up to reflection and scaling.