Talk:Ramanujan's master theorem

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Add a bottom section.

Hello, I would like to add a bottom section to enhance the Wikipedia page. The topic added is not large enough to merit a separate page. The additional content will enhance the quality of the Wikipedia page. The references overlap with the existing references. I added 2 additional references, added labels to the references, and capitalized Ramanujan's Master Theorem as it is a proper noun. Thanks.

Proposal

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.

Page from Ramanujan's notebook stating his Master theorem.

The result is stated as follows:

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

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

then the Mellin transform of ${\textstyle f(x)}$ is given by

$\int _{0}^{\infty }x^{s-1}f(x)dx=\Gamma (s)\,\varphi (-s)$

where ${\textstyle \Gamma (s)}$ 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 ${\textstyle \lambda (n)\equiv {\frac {\varphi (n)}{\,\Gamma (1+n)\,}}}$ and using the functional equation for the gamma function.

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

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 ${\textstyle B_{k}(x)}$ 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 ${\textstyle \zeta (1-n,a)=-{\frac {B_{n}(a)}{n}}}$ for ${\textstyle ~n\geq 1}$. 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 ${\textstyle 0<{\mathcal {Re}}(s)<1}$.

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 ${\textstyle \zeta (k)}$ 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 ${\textstyle 0<{\mathcal {Re}}(s)<1}$.

Special cases of ${\textstyle s={\frac {1}{2}}}$ and ${\textstyle s={\frac {3}{4}}}$ 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 ${\textstyle 0<2{\mathcal {Re}}(s)<{\mathcal {Re}}(\nu )+{\tfrac {3}{2}}}$.

Equivalently, if the spherical Bessel function ${\textstyle j_{\nu }(z)}$ 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 ${\textstyle 0<2{\mathcal {Re}}(s)<{\mathcal {Re}}(\nu )+2}$.

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

Bracket Integration Method

The Bracket Integration Method applies Ramanujan's Master Theorem to a broad range of integrals. The Bracket Integration Method generates an integral of a series expansion, introduces simplifying notations, solves linear equations, and completes the integration using formulas arising from Ramanujan's Master Theorem.

Generate an integral of a series expansion

This method transforms the integral to an integral of a series expansion involving M variables, ${\textstyle x_{1},\ldots x_{M}}$, and S summation parameters, ${\textstyle n_{1},\dots n_{S}}$. A multivariate integral may assume this form.

$\int _{0}^{\infty }...\int _{0}^{\infty }\sum _{n_{1},...,n_{S}=0}^{\infty }\varphi (n_{1}...n_{S})\ \prod _{j=1}^{S}\left({\frac {(-1)^{n_{j}}}{n_{j}!}}\right)\prod _{j=1}^{M}(x_{j})^{(-c_{j}+a_{j1}\cdot n_{1}+...+a_{jS}\cdot n_{S}-1)}\ dx_{1}...dx_{M}$ (B.0)

Apply special notations

• The bracket (${\textstyle <\dots >}$), indicator (${\textstyle \phi }$), and monomial power notations replace terms in the series expansion.

$\int _{0}^{\infty }x^{c+b\cdot n-1}\ dx\ \to \ <c+b\cdot n>$ (B.1)

${\frac {(-1)^{n}}{n!}}\ \to \ \phi _{n}$ (B.2)

$\prod _{j=1}^{S}\left({\frac {(-1)^{n_{j}}}{n_{j}!}}\right)\ \to \ \phi _{n_{1},...,n_{S}}$ (B.3)

$\left(\sum _{k=1}^{P}u_{k}\right)^{\mp d}\ \to \ \sum _{n_{1}...n_{P}=0}^{\infty }\phi _{n_{1}...n_{P}}\prod _{k=1}^{P}u_{k}^{n_{k}}{\frac {<\pm d+\sum _{j=1}^{P}n_{j}>}{\Gamma (\pm d)}}$ (B.4)

• Application of these notations transforms the integral to a bracket series containing B brackets.

$\sum _{n_{1},...,n_{S}=0}^{\infty }\varphi (n_{1}...n_{S})\ \phi (n_{1}...n_{S})\prod _{j=1}^{B}<-c_{j}+\sum _{k=1}^{S}a_{jk}\cdot n_{k}>$ (B.5)

• Each bracket series has an index defined as index=number of sums - number of brackets.
• Among all bracket series representations of an integral, the representation with a minimal index is preferred.

Solve linear equations

• The array of coefficients ${\textstyle a_{jk}}$ must have maximum rank, linearly independent leading columns to solve the following set of linear equations.
• If the index is non-negative, solve this equation set for each ${\textstyle n_{j}^{\ast }}$. The terms ${\textstyle n_{j}^{\ast }}$ may be linear functions of ${\textstyle \{n_{B+1}\dots n_{S}\}}$.

$-c_{j}+\sum _{k=1}^{B}a_{jk}\cdot n_{k}^{\ast }+\sum _{k=B+1}^{S}a_{jk}\cdot n_{k}=0$ (B.6)

• If the index is zero, equation (B.6) simplifies to solving this equation set for each ${\textstyle n_{j}^{\ast }}$

$-c_{j}+\sum _{k=1}^{B}a_{jk}\cdot n_{k}^{\ast }=0$ (B.7)

• If the index is negative, the integral cannot be determined.

Apply formulas

• If the index is non-negative, the formula for the integral is this form.

$\sum _{n_{B+1}\dots n_{S}=0}^{\infty }{\frac {\varphi (n_{1}^{\ast }\dots n_{B}^{\ast },n_{B+1}\dots n_{S})\cdot \prod _{j=1}^{B}\Gamma (-n_{j}^{\ast })}{det|A|}}$ (B.8)

• These rules apply.
  • A series is generated for each choice of free summation parameters, ${\textstyle \{n_{B}+1,\dots N_{S}\}}$.
  • Series converging in a common region are added.
  • If a choice generates a divergent series or null series (a series with zero valued terms), the series is rejected.
  • A bracket series of negative index is assigned no value.
  • If all series are rejected, then the method cannot be applied.
  • If the index is zero, the formula B.8 simplifies to this formula and no sum occurs.

$\ {\frac {\varphi (n_{1}^{\ast }\dots n_{S}^{\ast })\cdot \prod _{j=1}^{S}\Gamma (-n_{j}^{\ast })}{det|A|}}$ (B.9)

Mathematical Basis

• Apply this variable transformation to the general integral form (B.0).

$y_{k}=x_{1}^{a_{1k}}\cdot \ldots \cdot x_{M}^{a_{Mk}}$ (B.10)

.

• This is the transformed integral (B.11) and the result from applying Ramanujan's Master Theorem (B.12).

$(det|A|^{-1})\cdot \int _{0}^{\infty }...\int _{0}^{\infty }\sum _{n_{1},...,n_{S}=0}^{\infty }\varphi (n_{1}...n_{S})\prod _{j=1}^{S}\left({\frac {(-1)^{n_{j}}}{n_{j}!}}\right)\prod _{j=1}^{M}(y_{j})^{-n_{j}^{\ast }+n_{j}-1}\ dy_{1}...dy_{M}$ (B.11)

$=\sum _{n_{M+1}\dots n_{S}=0}^{\infty }{\frac {\varphi (n_{1}^{\ast }\dots n_{M}^{\ast },n_{M+1}\dots n_{S})\cdot \prod _{j=1}^{M}\Gamma (-n_{j}^{\ast })}{det|A|}}$ (B.12)

• The number of brackets (B) equals the number of integrals (M) (B.1). In addition to generating the algorithm's formulas (B.8,B.9), the variable transformation also generates the algorithm's linear equations (B.6,B.7).

Example

• The Bracket Integration Method is applied to this integral.

$\int _{0}^{\infty }x^{3/2}\cdot e^{-x^{3}/2}\ dx$

• Generate the integral of a series expansion (B.0).

$\int _{0}^{\infty }\sum _{n=0}^{\infty }2^{-n}\cdot {\frac {(-1)^{n}}{n!}}\cdot x^{(3\cdot n+5/2)-1}\ dx$

• Apply special notations (B.1, B.2).

$\sum _{n=0}^{\infty }2^{-n}\cdot \phi (n)\cdot <3\cdot n+{\frac {5}{2}}>$

• Solve the linear equation (B.7).

$3\cdot n^{\ast }+{\frac {5}{2}}=0,\ n^{\ast }={\frac {-5}{6}}$

• Apply the formula (B.9).

${\frac {2^{\frac {5}{6}}\cdot \Gamma ({\frac {5}{6}})}{3}}$

External links

• "Ramanujan's Master Theorem". mathworld.wolfram.com.
• "rmt" (PDF). ArminStraub. publications.

TMM53 (talk) 05:08, 7 October 2022 (UTC) TMM53 (talk) 05:08, 7 October 2022 (UTC)

Revision

This is a proposed revision. Although current content is correct, a rewrite makes the content easier to understand for the reader. Additional content and references were added. If any concerns, please send me a message, and let us discuss. ThanksTMM53 (talk) 05:59, 21 May 2024 (UTC)