Ramanujan's master theorem

In mathematics, Ramanujan's Master Theorem, named after Srinivasa Ramanujan, ^[1] 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

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

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

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

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. ^[2]

A similar result was also obtained by Glaisher. ^[3]

Alternative formalism

An alternative formulation of Ramanujan's Master Theorem is as follows:

$${\displaystyle \int _{0}^{\infty }x^{s-1}\left(\,\lambda (0)-x\,\lambda (1)+x^{2}\,\lambda (2)-\,\cdots \,\right)dx={\frac {\pi }{\,\sin(\pi s)\,}}\,\lambda (-s)}$$

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 ${\textstyle 0<\operatorname {\mathcal {Re}} (s)<1}$ subject to growth conditions on ${\textstyle \varphi }$. ^[4]

Proof

A proof subject to "natural" assumptions (though not the weakest necessary conditions) to Ramanujan's Master theorem was provided by G. H. Hardy ^[5] (chapter XI) 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:

$${\displaystyle {\frac {z\,e^{x\,z}}{\,e^{z}-1\,}}=\sum _{k=0}^{\infty }B_{k}(x)\,{\frac {z^{k}}{k!}}}$$

These polynomials are given in terms of the Hurwitz zeta function:

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

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: ^[6]

$${\displaystyle \int _{0}^{\infty }x^{s-1}\left({\frac {e^{-ax}}{\,1-e^{-x}\,}}-{\frac {1}{x}}\right)dx=\Gamma (s)\,\zeta (s,a)\!}$$

which is valid for ${\textstyle 0<\operatorname {\mathcal {Re}} (s)<1}$.

Application to the gamma function

Weierstrass's definition of the gamma function

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

is equivalent to expression

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

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

Then applying Ramanujan master theorem we have:

$${\displaystyle \int _{0}^{\infty }x^{s-1}{\frac {\,\gamma \,x+\log \Gamma (1+x)\,}{x^{2}}}\mathrm {d} x={\frac {\pi }{\sin(\pi s)}}{\frac {\zeta (2-s)}{2-s}}\!}$$

valid for ${\textstyle 0<\operatorname {\mathcal {Re}} (s)<1}$.

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

$${\displaystyle \int _{0}^{\infty }{\frac {\,\gamma x+\log \Gamma (1+x)\,}{x^{5/2}}}\,\mathrm {d} x={\frac {2\pi }{3}}\,\zeta \left({\frac {3}{2}}\right)}$$

$${\displaystyle \int _{0}^{\infty }{\frac {\,\gamma \,x+\log \Gamma (1+x)\,}{x^{9/4}}}\,\mathrm {d} x={\sqrt {2}}{\frac {4\pi }{5}}\zeta \left({\frac {5}{4}}\right)}$$

Application to Bessel functions

The Bessel function of the first kind has the power series

$${\displaystyle J_{\nu }(z)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\Gamma (k+\nu +1)k!}}{\bigg (}{\frac {z}{2}}{\bigg )}^{2k+\nu }}$$

By Ramanujan's Master Theorem, together with some identities for the gamma function and rearranging, we can evaluate the integral

$${\displaystyle {\frac {2^{\nu -2s}\pi }{\sin {(\pi (s-\nu ))}}}\int _{0}^{\infty }z^{s-1-\nu /2}J_{\nu }({\sqrt {z}})\,dz=\Gamma (s)\Gamma (s-\nu )}$$

valid for ${\textstyle 0<2\operatorname {\mathcal {Re}} (s)<\operatorname {\mathcal {Re}} (\nu )+{\tfrac {3}{2}}}$.

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

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

valid for ${\textstyle 0<2\operatorname {\mathcal {Re}} (s)<\operatorname {\mathcal {Re}} (\nu )+2}$.

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 ${\textstyle s={\frac {1}{2}}}$ or ${\textstyle s=1}$ gives the gamma function by itself, up to reflection and scaling.

Bracket integration method

The bracket integration method (method of brackets) applies Ramanujan's Master Theorem to a broad range of integrals. ^[7] The bracket integration method generates the integrand's series expansion, creates a bracket series, identifies the series coefficient and formula parameters and computes the integral. ^[8]

Integration formulas

This section identifies the integration formulas for integrand's with and without consecutive integer exponents and for single and double integrals. The integration formula for double integrals may be generalized to any multiple integral. In all cases, there is a parameter value ${\textstyle n^{\ast }}$ or array of parameter values ${\textstyle N^{\ast }}$ that solves one or more linear equations derived from the exponent terms of the integrand's series expansion.

Consecutive integer exponents, 1 variable

This is the function series expansion, integral and integration formula for an integral whose integrand's series expansion contains consecutive integer exponents. ^[9]

$${\displaystyle {\begin{aligned}&f(y)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ y^{n}\\&\int _{0}^{\infty }y^{c-1}f(y)\,dy\\&=\int _{0}^{\infty }\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ y^{n+c-1}dy\\&=\Gamma (-n^{\ast })\,\varphi (n^{\ast }).\end{aligned}}}$$

The parameter ${\displaystyle n^{\ast }}$ is a solution to this linear equation.

$${\displaystyle n^{\ast }+c=0,\ n^{\ast }=-c}$$

General exponents, 1 variable

Applying the substitution ${\textstyle y=x^{a}}$ generates the function series expansion, integral and integration formula for an integral whose integrand's series expansion may not contain consecutive integer exponents. ^[8]

$${\displaystyle {\begin{aligned}&f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ x^{an}\\&\int _{0}^{\infty }x^{c-1}f(x)\,dx\\&=\int _{0}^{\infty }\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ x^{an+c-1}dx\\&=a^{-1}\ \Gamma (-n^{\ast })\,\varphi (n^{\ast }).\\\end{aligned}}}$$

The parameter ${\textstyle n^{\ast }}$ is a solution to this linear equation.

$${\displaystyle a\ n^{\ast }+c=0,\ n^{\ast }=-a^{-1}c}$$

Consecutive integer exponents, double integral

This is the function series expansion, integral and integration formula for a double integral whose integrand's series expansion contains consecutive integer exponents. ^[10]

$${\displaystyle {\begin{aligned}&f(y_{1},y_{2})=\sum _{n=0}^{\infty }{\frac {(-1)^{n_{1}}}{n_{1}!}}{\frac {(-1)^{n_{2}}}{n_{2}!}}\ \varphi (n_{1},n_{2})\ y_{1}^{n_{1}}\ y_{2}^{n_{2}}\\&\int _{0}^{\infty }y_{1}^{c_{1}-1}y_{2}^{c_{2}-1}\ f(y_{1},y_{2})\ dy_{1}\ dy_{2}\\&=\int _{0}^{\infty }\int _{0}^{\infty }\sum _{n_{1}=0}^{\infty }\sum _{n_{2}=0}^{\infty }{\frac {(-1)^{n_{1}}}{n_{1}!}}{\frac {(-1)^{n_{2}}}{n_{2}!}}\ \varphi (n_{1},n_{2})\ y_{1}^{n_{1}+c_{1}-1}\ y_{2}^{n_{2}+c_{2}-1}\ dy_{1}\ dy_{2}\\&=\Gamma (-n_{1}^{\ast })\ \Gamma (-n_{2}^{\ast })\ \varphi (n_{1}^{\ast },n_{2}^{\ast }).\\\end{aligned}}}$$

The parameters ${\textstyle n_{1}^{\ast }}$ and ${\textstyle n_{2}^{\ast }}$ are solutions to these linear equations.

$${\displaystyle n_{1}^{\ast }+c_{1}=0,\ n_{2}^{\ast }+c_{2}=0,\ n_{1}^{\ast }=-c_{1},\ n_{2}^{\ast }=-c_{2}}$$

General exponents, double integral

This section describes the integration formula for a double integral whose integrand's series expansion may not contain consecutive integer exponents. Matrices contain the parameters needed to express the exponents in a series expansion of the integrand, and the determinant of invertible matrix ${\textstyle A}$ is ${\textstyle \operatorname {det} |A|}$. ^[11]

$${\displaystyle A={\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}},\ C={\begin{vmatrix}c_{1}\\c_{2}\end{vmatrix}},\ \ N^{\ast }={\begin{vmatrix}n_{1}^{\ast }\\n_{2}^{\ast }\end{vmatrix}}}$$

Substitute variables ${\textstyle y_{1},y_{2}}$ with variables ${\textstyle x_{1},x_{2}}$ to generate an integrand's series expansion whose exponents may not be consecutive integers. ^[10]

$${\displaystyle y_{1}=x_{1}^{a_{11}}x_{2}^{a_{21}},\quad y_{2}=x_{1}^{a_{12}}x_{2}^{a_{22}}}$$

This is the integral and integration formula. ^{[12] [13]}

$${\displaystyle {\begin{aligned}&\int _{0}^{\infty }\int _{0}^{\infty }\sum _{n_{1}=0}^{\infty }\sum _{n_{2}=0}^{\infty }{\frac {(-1)^{n_{1}}}{n_{1}!}}{\frac {(-1)^{n_{2}}}{n_{2}!}}\ \varphi (n_{1},n_{2})\ x_{1}^{n_{1}a_{11}+n_{2}a_{12}+c_{1}-1}\ x_{2}^{n_{1}a_{21}+n_{2}a_{22}+c_{2}-1}\ dx_{1}\ dx_{2}\\&=\operatorname {det} |A|^{-1}\ \Gamma (-n_{1}^{\ast })\ \Gamma (-n_{2}^{\ast })\ \varphi (n_{1}^{\ast },n_{2}^{\ast })\\\end{aligned}}}$$

The parameter matrix ${\textstyle N^{\ast }}$ is a solution to this linear equation. ^[14]

$${\displaystyle A\ N^{\ast }+C=0,\quad N^{\ast }=-A^{-1}C}$$

.

Positive complexity index

In some cases, there may be more sums then variables. For example, if the integrand is a product of 3 functions of a common single variable, and each function is converted to a series expansion sum, the integrand is now a product of 3 sums, each sum corresponding to a distinct series expansion.

• The number of brackets is the number of linear equations associated with an integral. This term reflects the common practice of bracketing each linear equation. ^[15]
• The complexity index is the number of integrand sums minus the number of brackets (linear equations). Each series expansion of the integrand contributes one sum. ^[15]
• The summation indices (variables) are the indices that index terms in a series expansion. In the example, there are 3 summation indices ${\textstyle n_{1},n_{2}}$ and ${\textstyle n_{3}}$ because the integrand is a product of 3 series expansions. ^[16]
• The free summation indices (variables) are the summation indices that remain after completing all integrations. Integration reduces the number of sums in the integrand by replacing the series expansions (sums) with an integration formula. Therefore, there are fewer summation indices after integration. The number of chosen free summation indices equals the complexity index. ^[16]

Integrals with a positive complexity index

The set of selected free indices, ${\textstyle F}$, contains indices ${\textstyle {\bar {n}}_{1},\ldots ,{\bar {n}}_{f}}$. Matrices ${\textstyle {\bar {A}}}$ and ${\textstyle {\bar {N}}}$ contain matrix elements that multiply free summation indices.

$${\displaystyle {\bar {A}}={\begin{vmatrix}{\bar {a}}_{11}&\ldots &{\bar {a}}_{1f}\\\vdots &&\vdots \\{\bar {a}}_{b1}&\ldots &{\bar {a}}_{bf}\end{vmatrix}},\ {\bar {N}}={\begin{vmatrix}{\bar {n}}_{1}\\\vdots \\{\bar {n}}_{f}\end{vmatrix}}}$$

.

The remaining indices are set ${\textstyle B}$ containing indices ${\textstyle n_{1},\ldots ,n_{b}}$. Matrices ${\textstyle A,C}$ and ${\textstyle N^{\ast }}$ contain matrix elements that multiply or sum with the remaining indices. The selected indices are chosen to make matrix ${\textstyle A}$ non-singular.

$${\displaystyle A={\begin{vmatrix}a_{11}&\ldots &a_{1b}\\\vdots &&\vdots \\a_{b1}&\ldots &a_{bb}\end{vmatrix}},\ C={\begin{vmatrix}c_{1}\\\vdots \\c_{b}\end{vmatrix}},\ \ N^{\ast }={\begin{vmatrix}n_{1}^{\ast }\\\vdots \\n_{b}^{\ast }\end{vmatrix}}}$$

.

This is the function's series expansion, integral and integration formula. ^[17]

$${\displaystyle {\begin{aligned}&f(x_{1},\ldots ,x_{b})\\&=\sum _{n\in B}^{\infty }\sum _{{\bar {n}}\in F}^{\infty }{\frac {(-1)^{n_{1}}}{n_{1}!}}{\frac {(-1)^{{\bar {n}}_{1}}}{{\bar {n}}_{1}!}}\ldots {\frac {(-1)^{n_{b}}}{n_{b}!}}{\frac {(-1)^{\bar {n_{f}}}}{{\bar {n}}_{f}!}}\varphi (n_{1},\ldots ,n_{b},{\bar {n}}_{1},\dots ,{\bar {n}}_{f})\prod _{x_{k}\in B}x_{k}^{n_{1}a_{k1}+\dots +{\bar {n}}_{1}{\bar {a}}_{k1}+\dots +c_{k}-1}\\&\int _{0}^{\infty }\ldots \int _{0}^{\infty }x^{c_{1}-1}\dots x^{c_{b}-1}f(x_{1},\ldots ,x_{b})\ dx_{1}\ldots dx_{b}\\&=\operatorname {det} |A|^{-1}\sum _{{\bar {n}}\in F}^{\infty }{\frac {(-1)^{{\bar {n}}_{1}}}{{\bar {n}}_{1}!}}\ldots {\frac {(-1)^{{\bar {n}}_{f}}}{{\bar {n}}_{f}!}}\ \Gamma (-n_{1}^{\ast })\ldots \Gamma (-n_{b}^{\ast })\ \varphi (n_{1}^{\ast },\ldots ,n_{b}^{\ast },{\bar {n}}_{1},\dots ,{\bar {n}}_{f}).\end{aligned}}}$$

The parameters ${\textstyle n_{1}^{\ast },\ldots ,n_{b}^{\ast }}$ are linear functions of the parameters ${\textstyle {\bar {n}}_{1}^{\ast },\ldots ,{\bar {n}}_{f}^{\ast }}$. ^[18]

$${\displaystyle A\ N^{\ast }+{\bar {A}}\ {\bar {N}}+C=0,\ N^{\ast }=-A^{-1}({\bar {A}}\ {\bar {N}}+C)}$$

.

Algorithm

Generate the integrand's series expansion

This may be a single series expansion or a product of series expansions.

Create a bracket series

A bracket series arises by substituting simplified bracket notations for terms of the integrand's series expansion.

• The indicator notation, ${\textstyle \phi }$, replace these terms. ^[19]

$${\displaystyle {\frac {(-1)^{n}}{n!}}\ \to \ \phi _{n}}$$

$${\displaystyle \prod _{j=1}^{N}\left({\frac {(-1)^{n_{j}}}{n_{j}!}}\right)\ \to \ \phi _{n_{1},\ldots ,n_{N}}}$$

.

• The bracket notation, ${\textstyle \langle \cdots \rangle }$, replaces these terms.

$${\displaystyle \int _{0}^{\infty }x^{c+a\ n-1}\ dx\ \to \langle c+a\ n\rangle }$$

• The reciprocal of a sum of variables raised to a positive power, ${\displaystyle \alpha }$, may be replaced by this bracket series. ^[4]

$${\displaystyle {\frac {1}{(x_{1}+\ldots +x_{b})^{\alpha }}}}$$

$${\displaystyle \to \sum _{m_{1}}\ldots \sum _{m_{b}}\ \phi _{m_{1},\dots ,m_{b}}\ x^{m_{1}}\dots x^{m_{b}}{\frac {\langle \alpha +m_{1}+\ldots +m_{b}\rangle }{\Gamma (\alpha )}}}$$

• This example transforms a series expansion to a bracket series.

$${\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }\sum _{n_{1}=0}^{\infty }\sum _{n_{2}=0}^{\infty }{\frac {(-1)^{n_{1}}}{n_{1}!}}{\frac {(-1)^{n_{2}}}{n_{2}!}}\ \varphi (n_{1},n_{2})\ x_{1}^{n_{1}a_{11}+n_{2}a_{12}+c_{1}-1}\ x_{2}^{n_{1}a_{21}+n_{2}a_{22}+c_{2}-1}\ dx_{1}\ dx_{2}}$$

$${\displaystyle \to \sum _{n_{1}=0}^{\infty }\sum _{n_{2}=0}^{\infty }\phi _{n_{1},n_{2}}\varphi (n_{1},n_{2})\langle n_{1}a_{11}+n_{2}a_{12}+c_{1}\rangle \langle n_{1}a_{21}+n_{2}a_{22}+c_{2}\rangle }$$

Identify the linear equation(s)

Set the expressions in brackets equal to zero. For multiple integrals, this generates equations involving matrices ${\textstyle A,N^{\ast }}$ and ${\textstyle C}$. For a positive complexity index bracket series, generate matrices ${\textstyle {\bar {A}}}$ and ${\textstyle {\bar {N}}}$.

Identify the series coefficient and parameters

Inspect the bracket series and identify the series coefficient ${\textstyle \varphi ()}$ and parameter ${\textstyle a}$ (single integral) or ${\textstyle \operatorname {det} |A|}$ (multiple integrals). Solve the linear equations to determine the parameter ${\textstyle n^{\ast }}$ (single integral) or parameter matrix ${\textstyle N^{\ast }}$ (multiple integrals).

Compute integrals

• If the complexity index is negative, more brackets than sums, the bracket series is assigned no value. ^[8]
• If the complexity index is zero, this is the function series expansion, single integral and integration formula. ^[9]

$${\displaystyle {\begin{aligned}&f(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ x^{an}\\&\int _{0}^{\infty }x^{c-1}f(x)\,dx\\&=\int _{0}^{\infty }\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\ \varphi (n)\ x^{an+c-1}dx\\&=a^{-1}\ \Gamma (-n^{\ast })\,\varphi (n^{\ast })\quad n^{\ast }=-c/a\\\end{aligned}}}$$

• If the complexity index is zero, this is the function series expansion, multiple integral and integration formula. ^[8]

$${\displaystyle \int _{0}^{\infty }\ldots \int _{0}^{\infty }x^{c_{1}-1}\dots x^{c_{b}-1}f(x_{1},\ldots ,x_{b})\ dx_{1}\ldots dx_{b}}$$

$${\displaystyle =\operatorname {det} |A|^{-1}\ \Gamma (-n_{1}^{\ast })\ldots \Gamma (-n_{b}^{\ast })\ \varphi (n_{1}^{\ast },\ldots ,n_{b}^{\ast })\ }$$

$${\displaystyle N^{\ast }=-A^{-1}C}$$

• If the complexity index is positive, this is the integral and integration formula. ^[20]

$${\displaystyle \int _{0}^{\infty }\ldots \int _{0}^{\infty }x^{c_{1}-1}\dots x^{c_{b}-1}f(x_{1},\ldots ,x_{b})\ dx_{1}\ldots dx_{b}}$$

$${\displaystyle =\operatorname {det} |A|^{-1}\sum _{{\bar {n}}\in F}^{\infty }{\frac {(-1)^{{\bar {n}}_{1}}}{{\bar {n}}_{1}!}}\ldots {\frac {(-1)^{{\bar {n}}_{f}}}{{\bar {n}}_{f}!}}\ \Gamma (-n_{1}^{\ast })\ldots \Gamma (-n_{b}^{\ast })\ \varphi (n_{1}^{\ast },\ldots ,n_{b}^{\ast },{\bar {n}}_{1},\dots ,{\bar {n}}_{f})}$$

$${\displaystyle N^{\ast }=-A^{-1}({\bar {A}}\ {\bar {N}}+C)}$$

.

For a positive complexity index bracket series, apply these rules. ^[8]

• Generate a series for each possible choice of free summation indices.
• Among all bracket series representations of an integral, the representation with a minimal complexity index is preferred.
• If a choice generates a divergent series or null series (a series with zero valued terms), the series is rejected.
• If all series are rejected, then the method cannot be applied.
• Series converging in a common region are added.

Example

Zero complexity index

The bracket method will integrate this integral.

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

• Generate the integrand's series expansion.

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

• Create a bracket series.

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

• This is the linear equation.

$${\displaystyle 3\ n^{\ast }+{\frac {5}{2}}=0}$$

• There is 1 bracket and 1 sum so the complexity index is zero. Identify the series coefficient and parameters.

$${\displaystyle \varphi (n)=2^{-n}}$$

$${\displaystyle a=3,\ c=5/2,\ n^{\ast }=-c/a=-5/6}$$

• Compute the integral.

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

$${\displaystyle =a^{-1}\ \Gamma (-n^{\ast })\ \varphi (n^{\ast })}$$

$${\displaystyle ={\frac {\Gamma \left({\frac {5}{6}}\right)\ 2^{5/6}}{3}}}$$

Positive complexity index

The bracket method will integrate this integral.

$${\displaystyle \int _{0}^{\infty }{\frac {1}{(1+x^{3}+x^{5})^{1/2}}}\ dx}$$

Substitute the bracket series for the reciprocal of a sum raised to a power.

$${\displaystyle \int _{0}^{\infty }\sum _{n_{1},n_{2},n_{3}}\ {\frac {1}{\sqrt {\Gamma (1/2)}}}\phi _{123}1^{n_{1}}x^{5n_{2}+3n_{3}}\langle n_{1}+n_{2}+n_{3}+1/2\rangle \ dx}$$

Apply the bracket notation.

$${\displaystyle \int _{0}^{\infty }\sum _{n_{1},n_{n},n_{3}}{\frac {1}{\sqrt {\Gamma (1/2)}}}\phi _{123}\langle 5\ n_{2}+3\ n_{3}+1\rangle \langle n_{1}+n_{2}+n_{3}+1/2\rangle }$$

There are 3 sums and 2 brackets; therefore, the complexity index is 1, and 1 index must be selected as a free index. Select ${\textstyle n_{3}}$ as the free index, ${\textstyle {\bar {n}}_{3}}$, and identify the linear equations.

$${\displaystyle 5n_{2}^{\ast }+3{\bar {n}}_{3}+1=0,\ n_{1}^{\ast }+n_{2}^{\ast }+{\bar {n}}_{3}+1/2=0}$$

$${\displaystyle AN^{\ast }+{\bar {A}}{\bar {N}}+C=0}$$

$${\displaystyle {\begin{vmatrix}1&1\\0&5\end{vmatrix}}{\begin{vmatrix}n_{1}^{\ast }\\n_{2}^{\ast }\end{vmatrix}}+{\begin{vmatrix}1\\3\end{vmatrix}}{\begin{vmatrix}{\bar {n}}_{3}\end{vmatrix}}+{\begin{vmatrix}1/2\\1\end{vmatrix}}=0}$$

Identify series coefficient function and parameters

$${\displaystyle \varphi (n_{1}^{\ast },n_{2}^{\ast },{\bar {n}}_{3})={\frac {1}{\sqrt {\Gamma (1/2)}}}={\frac {1}{\sqrt {\pi }}}}$$

$${\displaystyle \operatorname {det} |A|=5}$$

$${\displaystyle n_{1}^{\ast }=-{\frac {2}{5}}{\bar {n}}_{3}-{\frac {3}{10}},\ n_{2}^{\ast }=-{\frac {3}{5}}{\bar {n}}_{3}-{\frac {1}{5}}}$$

Compute the integral.

$${\displaystyle {\begin{aligned}&\int _{0}^{\infty }{\frac {1}{(1+x^{3}+x^{5})^{1/2}}}\ dx\\&=\sum _{{\bar {n}}_{3}=0}^{\infty }{\frac {(-1)^{{\bar {n}}_{3}}}{{\bar {n}}_{3}!}}\operatorname {det} |A|^{-1}\Gamma (-n_{1}^{\ast })\Gamma (-n_{2}^{\ast })\varphi (n_{1}^{\ast },n_{2}^{\ast },{\bar {n}}_{3})\\&=\sum _{{\bar {n}}_{3}=0}^{\infty }{\frac {(-1)^{{\bar {n}}_{3}}}{{\bar {n}}_{3}!}}{\frac {\Gamma ({\frac {2}{5}}{\bar {n}}_{3}+{\frac {3}{10}})\Gamma ({\frac {3}{5}}{\bar {n}}_{3}+{\frac {1}{5}})}{5{\sqrt {\pi }}}}\end{aligned}}}$$

Citations

1. Berndt 1985.
2. González, Moll & Schmidt 2011.
3. Glaisher 1874, pp. 53–55.
4. Amdeberhan et al. 2012, pp. 103–120.
5. Hardy 1978.
6. Espinosa & Moll 2002, pp. 449–468. sfn error: no target: CITEREFEspinosaMoll2002 (help)
7. Gonzalez & Moll 2010, pp. 50–73.
8. Gonzalez, Jiu & Moll 2020, pp. 983–985.
9. Amdeberhan et al. 2012, p. 117, Eqn. 9.5.
10. Amdeberhan et al. 2012, p. 118.
11. Ananthanarayan et al. 2023, Eqn. 7.
12. Amdeberhan et al. 2012, p. 118, Eqn. 9.6.
13. Ananthanarayan et al. 2023, Eqn. 8.
14. Ananthanarayan et al. 2023, Eqn. 9.
15. Gonzalez et al. 2022, p. 28.
16. Amdeberhan et al. 2012, p. 117.
17. Ananthanarayan et al. 2023, Eqn. 10.
18. Ananthanarayan et al. 2023, Eqn. 11.
19. González, Moll & Schmidt 2011, p. 8.
20. Ananthanarayan et al. 2023, Eqns. 9-11.

References

• Amdeberhan, Tewodros; Gonzalez, Ivan; Harrison, Marshall; Moll, Victor H.; Straub, Armin (2012). "Ramanujan's Master Theorem". The Ramanujan Journal. 29 (1–3): 103–120. CiteSeerX   10.1.1.232.8448 . doi:10.1007/s11139-011-9333-y. S2CID   8886049.
• Ananthanarayan, B.; Banik, Sumit; Friot, Samuel; Pathak, Tanay (2023). "Method of Brackets: Revisiting a technique for calculating Feynman integrals and certain definite integrals". Physical Review D. 108 (8).
• Berndt, B. (1985). Ramanujan's Notebooks, Part I. New York: Springer-Verlag.
• Glaisher, J.W.L. (1874). "A new formula in definite integrals". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 48 (315): 53–55. doi:10.1080/14786447408641072.
• González, Iván; Moll, V.H.; Schmidt, Iván (2011). "A generalized Ramanujan Master Theorem applied to the evaluation of Feynman diagrams". arXiv: 1103.0588 [math-ph].
• Gonzalez, Ivan; Moll, Victor H. (2010). "Definite integrals by the method of brackets. Part 1,". Advances in Applied Mathematics. 45 (1): 50–73. doi:10.1016/j.aam.2009.11.003.
• Gonzalez, Ivan; Kohl, Karen; Jiu, Lin; Moll, Victor H. (1 January 2017). "An extension of the method of brackets. Part 1". Open Mathematics. 15 (1): 1181–1211. doi:10.1515/math-2017-0100. ISSN   2391-5455.
• Gonzalez, Ivan; Kohl, Karen; Jiu, Lin; Moll, Victor H. (March 2018). "The Method of Brackets in Experimental Mathematics". Frontiers in Orthogonal Polynomials and q -Series. WORLD SCIENTIFIC. pp. 307–318. ISBN   978-981-322-887-0.
• Gonzalez, Ivan; Jiu, Lin; Moll, Victor H. (2020). "An extension of the method of brackets. Part 2". Open Mathematics. 18 (1): 983–995. arXiv: 1707.08942 . doi: 10.1515/math-2020-0062 . ISSN   2391-5455.
• Gonzalez, Ivan; Kondrashuk, Igor; Moll, Victor H.; Recabarren, Luis M. (2022). "Mellin–Barnes integrals and the method of brackets". The European Physical Journal C. 82 (1): 28. doi:10.1140/epjc/s10052-021-09977-x. ISSN   1434-6052.
• Hardy, G.H. (1978). Ramanujan: Twelve lectures on subjects suggested by his life and work (3rd ed.). New York, NY: Chelsea. ISBN   978-0-8284-0136-4.

External links

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