Proving Ramanujan's Integral Formula

Question

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?

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

Answer 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 !

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

Answer 2

\begin{align*} f(z) &= \prod_{\ell = 1}^{n} \left((b + \ell)^{2} + z^{2}\right), \quad g(z) = \prod_{\ell = 0}^{n} \left((a + \ell)^{2} + z^{2}\right) \\ \gamma_{R} &\text{ is the half-moon in the upper complex half-plane.} \\ f(i(a+k)) &= \prod_{\ell = 1}^{n} \left((b + \ell)^{2} - (a+k)^{2}\right) = \frac{(b+a+k+n)!}{(b+a+k)!} \cdot \frac{(b-a-k+n)!}{(b-a-k)!} \\ g'(i(a+k)) &= (-1)^{k} \cdot k! \cdot \frac{(2a+2k-1)!}{(2a+k-1)!} \cdot 2i(a+k) \cdot (n-k)! \cdot \frac{(2a+k+n)!}{(2a+2k)!} \\ &= i \cdot (-1)^{k} \cdot \frac{(2a+k+n)!}{(2a+k-1)!} \cdot k! \cdot (n-k)! \\ \int_{-\infty}^{\infty} &\frac{1}{a^{2}+x^{2}} \cdot \frac{(b+1)^{2}+x^{2}}{(a+1)^{2}+x^{2}} \cdots \frac{(b+n)^{2}+x^{2}}{(a+n)^{2}+x^{2}} \,dx \\ &= \lim_{R \to \infty} \oint_{\gamma_{R}} \frac{f(z)}{g(z)} \,dz = 2\pi i \sum_{k=0}^{n} \text{res}\left(\frac{f}{g}, i(a+k)\right) \\ &= 2\pi i \sum_{k=0}^{n} \frac{f\left(i(a+k)\right)}{g'\left(i(a+k)\right)} \\ &= 2\pi \sum_{k=0}^{n} (-1)^{k} \frac{(2a+k-1)!}{(2a+k+n)!} \frac{1}{k! (n-k)!} \frac{(b+a+k+n)! (b-a-k+n)!}{(b+a+k)! (b-a-k)!}. \end{align*}

If we multiply both sides by $a^{2}(a+1)^{2}\cdots(a+n)^{2} = \frac{(a+n)!^{2}}{(a-1)!^{2}}$ and divide by $(b+1)^{2}(b+2)^{2}\cdots(b+n)^{2} = \frac{(b+n)!^{2}}{b!^{2}}$, then \begin{align*} I_{n} &= 2\pi \frac{b!^{2}}{(a-1)!^{2}} \sum_{k=0}^{n} \underbrace{(-1)^{k} \frac{(2a+k-1)!}{k!} \frac{1}{(b+a+k)! (-2a-k)! (b-a-k)!}}_{A_{n,k}} \\ &\quad \times \underbrace{\frac{(b+a+k+n)! (b-a-k+n)!}{(2a+k+n)! (n-k)!} \frac{(a+n)!^{2}}{(b+n)!^{2}}}_{\text{denoted as } A_{n,k}}. \end{align*}

For $0 \leq k < M$, as $n \to \infty$, $\lim_{n \to \infty} A_{n,k} = 1$. Therefore, \begin{align*} I &= \lim_{n \to \infty} I_{n} \\ &= 2\pi \frac{b!^{2}(-2a)!(2a-1)!}{(a-1)!^{2}} \left(\sum_{k=0}^{M-1} \frac{A_{n,k}}{k! (b+a+k)! (-2a-k)! (b-a-k)!} + \sum_{k=M}^{n} \frac{A_{n,k}}{k! (b+a+k)! (-2a-k)! (b-a-k)!}\right) \\ &= 2\pi \frac{b!^{2}(-2a)!(2a-1)!}{(a-1)!^{2}} \left(\sum_{k=0}^{M-1} \frac{1}{k! (b+a+k)! (-2a-k)! (b-a-k)!} + \varepsilon_{M}\right), \end{align*} where $\varepsilon_{M}$ tends to zero as $M \to \infty$.

If $M \to \infty$, $\varepsilon_{M}$ tends to zero, and the hypergeometric series $\sum_{k=0}^{\infty} \frac{1}{k! (b+a+k)! (-2a-k)! (b-a-k)!}$ can be expressed as $\frac{(2b-2a)!}{(-2a)!(b-a)!(b-a)!(2b)!}$ for $2b-2a > -1$, or $a < b+\frac{1}{2}$. Therefore, [ I = 2\pi \frac{b!^{2}(2a-1)!(2b-2a)!}{(a-1)!^{2}(b-a)!^{2}(2b)!}. ]

This can be written using Legendre's duplication formula as \begin{align*} I &= \sqrt{\pi} \frac{\Gamma\left(a+\frac{1}{2}\right) \Gamma(b+1) \Gamma\left(b-a+\frac{1}{2}\right)}{\Gamma(a) \Gamma\left(b+\frac{1}{2}\right) \Gamma\left(b-a+1\right)}. \end{align*}

the integral (I) is expressed using Legendre's duplication formula as the square root of pi multiplied by the ratio of gamma functions: $$I = \sqrt{\pi} \frac{\Gamma\left(a+\frac{1}{2}\right) \Gamma(b+1) \Gamma\left(b-a+\frac{1}{2}\right)}{\Gamma(a) \Gamma\left(b+\frac{1}{2}\right) \Gamma\left(b-a+1\right)}$$