Proving Ramanujan's Integral Formula

In a letter to Hardy, Ramanujan described a simple identity valid for $0<a<b+\frac 12$:

\begin{align} & \small\int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx \\[5mm] = &\ \dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}\tag1 \end{align}

Which I find remarkable.

Questions:

1. Has anyone discovered a way to prove $(1)$? If so, how do you prove it?
2. Where did Ramanujan learn all of his integrational-calculus material (It doesn't appear in the Synopsis book)?
3. Does anyone know a pdf or book where I can start learning advanced integration?

I'm wondering how you would prove $(1)$ and if there are similar identities that can be made. Wikipedia doesn't have any information.

Solution 1:

Proposition 1 :$$\color{blue}{\displaystyle \sum\limits_{k=1}^\infty \dfrac{(-1)^{k-1}}{k}\;\zeta_H(k,a)x^k \; =\; \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\tag1}$$

Proof :

$$\begin{align*} & \displaystyle \sum\limits_{k=0}^{\infty}(-x)^{k}\zeta_H(k+1,a)\\ & \displaystyle = \sum\limits_{k,n=0}^{\infty} \dfrac{(-x)^k}{(n+a)^{k+1}}\\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a}\sum\limits_{k=0}^\infty \left(\dfrac{-x}{n+a}\right)^k \\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a+x} \\ & \displaystyle = -\psi(a+x)\end{align*}$$

Now integrating,

$$\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k =\ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\\ $$

Proposition 2 :$$\color{blue}{\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}\tag 2} $$

Proof :

It is sufficient to evaluate the series,

$$\begin{align*} & \displaystyle\sum\limits_{n=0}^{\infty}\ln\left(1+\dfrac{x}{n+a}\right) \\ & = \displaystyle \sum\limits_{k=1}^\infty\sum\limits_{n=0}^\infty \dfrac{(-1)^{k-1}}{k}x^k \dfrac{1}{(n+a)^k} \\ & =\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k \\ & =\displaystyle \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\end{align*}$$

$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x}{n+a}\right) = \dfrac{\Gamma(a)}{\Gamma(a+x)}$$

Therefore,

$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}$$

Proposition 3 : If $\; \displaystyle F(s)=\int\limits_0^\infty x^{s-1}f(x)\; dx\; $ then $$\color{blue}{\displaystyle \int\limits_{-\infty}^{\infty} |F(ix)|^2 \; dx \;= \; 2\pi\int\limits_0^\infty \dfrac{|f(x)|^2}{x}\; dx\tag 3} $$

Proof :

$$\displaystyle F(it)=\int\limits_0^\infty x^{it}\dfrac{f(x)}{x}\; dx$$

Set $x=e^y$ ,

$$\displaystyle F(it)=\int\limits_0^\infty e^{ixt}f(e^x)\; dx$$

Now by properties of Fourier Transform,

$$\begin{align*} & \displaystyle f(e^t)=\int\limits_{-\infty}^\infty g(x)e^{-ixt}\; dx \\ & \displaystyle g(t)=\dfrac{1}{2\pi}\int\limits_{-\infty}^\infty f(e^x)e^{ixt}\; dx\\\end{align*}$$

$$\begin{align*}\displaystyle F(it) & =2\pi g(t) \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 4\pi^2 \int\limits_{-\infty}^\infty |g(t)|^2\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty g(t) \int\limits_{-\infty}^\infty e^{ixt}f(e^x)\; dx\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x) \int\limits_{-\infty}^\infty e^{ixt}g(t)\; dt\; dx \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x)\overline{f(e^x)}\; dx =2\pi\int\limits_{-\infty}^\infty |f(e^x)|^2\; dx\end{align*}$$

Now by setting $e^x=t$ we get our result,

$$\displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt = 2\pi\int\limits_{-\infty}^\infty \dfrac{|f(t)|^2}{t}\; dt$$

Main Problem: $$\color{blue}{\displaystyle \int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx=\dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}} $$

Proof : If we denote the integral by $I$ then using $(2)$ it can be rewritten as,

$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{-\infty}^\infty \dfrac{|\Gamma(a+ix)|^2}{|\Gamma(b+1+ix)|^2}\; dx$$

Now by defining $\displaystyle h(x)=\dfrac{x^a (1-x)^{b-a}}{\Gamma(b-a+1)}$ for $x\in[0,1]$ and $0$ for $\forall x\notin[0,1]$ we can conclude that $\displaystyle F(s)=M[h(x)]=\dfrac{\Gamma(s+a)}{\Gamma(s+b+1)}$ and from $(3)$ it follows that,

$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{0}^1 \dfrac{|h(x)|^2}{x}\; dx = \dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}$$

where last line follows from the Duplication formula , and we are done !

$$\large \color{red}{\color{blue}{\boxed{\mathfrak{PROVED}}}} $$