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

File:Ramanujanmasterth.jpg

Page from Ramanujan's notebook stating his Master theorem.

The result is stated as follows:

Assume function $f(x)$ has an expansion of the form

$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{\varphi (k)}{k!}(-x{)}^k$

then Mellin transform of $f(x)$ is given by

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{s-1}f(x)dx=\mathrm{\Gamma }(s)\varphi (-s)$

where $\mathrm{\Gamma }(s)$ is the Gamma function.

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

Multidimensional version of this theorem also appear in quantum physics (through Feynman diagrams).

A similar result was also obtained by J. W. L. Glaisher.

Alternative formalism

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{s-1}(\lambda (0)-x\lambda (1)+x^2\lambda (2)-\cdots )dx=\frac{\pi }{\sin (\pi s)}\lambda (-s)$

which gets converted to original form after substituting $\lambda (n)=\frac{\varphi (n)}{\mathrm{\Gamma }(1+n)}$ and using functional equation for Gamma function.

The integral above is convergent for $0<\mathrm{\Re }(s)<1$.

Proof

The proof of Ramanujan's Master Theorem provided by G. H. Hardy employs Cauchy's residue theorem as well as the well-known Mellin inversion theorem.

Application to Bernoulli polynomials

The generating function of the Bernoulli polynomials $B_k(x)$ is given by:

$\frac{z{e}^{xz}}{e^z-1}=\sum _{k=0}^{\mathrm{\infty }}{B}_k(x)\frac{z^k}{k!}$

These polynomials are given in terms of Hurwitz zeta function:

$\zeta (s,a)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+a{)}^s}$

by $\zeta (1-n,a)=-\frac{B_n(a)}{n}$ for $n\ge 1$. By means of Ramanujan master theorem and generating function of Bernoulli polynomials one will have following integral representation:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{s-1}(\frac{e^{-ax}}{1-e^{-x}}-\frac{1}{x})dx=\mathrm{\Gamma }(s)\zeta (s,a)$

valid for $0<Re(s)<1$.

Application to the Gamma function

Weierstrass's definition of the Gamma function

$\mathrm{\Gamma }(x)=\frac{e^{-\gamma x}}{x}\prod _{n=1}^{\mathrm{\infty }}{(1+\frac{x}{n})}^{-1}{e}^{x/n}$

is equivalent to expression

$\log \mathrm{\Gamma }(1+x)=-\gamma x+\sum _{k=2}^{\mathrm{\infty }}\frac{\zeta (k)}{k}(-x{)}^k$

where $\zeta (k)$ is the Riemann zeta function.

Then applying Ramanujan master theorem we have:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{s-1}\frac{\gamma x+\log \mathrm{\Gamma }(1+x)}{x^2}dx=\frac{\pi }{\sin (\pi s)}\frac{\zeta (2-s)}{2-s}$

valid for $0<Re(s)<1$.

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\gamma x+\log \mathrm{\Gamma }(1+x)}{x^{5/2}}dx=\frac{2\pi }{3}\zeta \left(\frac{3}{2}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\gamma x+\log \mathrm{\Gamma }(1+x)}{x^{5/4}}dx=\sqrt{2}\frac{4\pi }{5}\zeta \left(\frac{5}{4}\right)$

Mathematica 7 is unable to compute these examples.

Evaluation of quartic integral

It is well known for the evaluation of

$F(a,m)=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{(x^4+2a{x}^2+1{)}^{m+1}}$

which is a well known quartic integral.

References

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External links

1. http://mathworld.wolfram.com/RamanujansMasterTheorem.html
2. http://www.youtube.com/watch?v=gLp5OsfUlNE
3. http://arminstraub.com/files/publications/rmt.pdf