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

$$\small\int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx=\dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}\tag1$$

Which I find remarkable.


  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.

  • 1
    $\begingroup$ A 'simple identity'... $\endgroup$ – Leon Sot Jan 31 '17 at 6:55
  • 2
    $\begingroup$ See mathoverflow.net/questions/66812/… $\endgroup$ – S.C.B. Jan 31 '17 at 8:06
  • $\begingroup$ I expect Ramanujan read some other math book(s) besides the Synopsis mentioned by Hardy. Likely at a rate of at least 2 lines per second. $\endgroup$ – DanielWainfleet Jan 31 '17 at 13:41
  • $\begingroup$ maybe one can show that the left hand and the right hand side have the same set of zeros and poles. this should fix the identity up to some constant which is hopefully not to difficult to obtain (this is more or less Liouvilles theorem if i remember correctly) $\endgroup$ – tired Jan 31 '17 at 18:08
  • 1
    $\begingroup$ Try the product formula for $\sin(x)$ and use some complex analysis / contour integrals $\endgroup$ – reuns Feb 1 '17 at 5:33

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,

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

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


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

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 $\forall x\notin[0,1]$ (Just like done in the link in the comment) 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 !

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

  • $\begingroup$ @Crescendo It's the Hurwitz Zeta function $\endgroup$ – Aditya Narayan Sharma Feb 28 '17 at 8:30
  • $\begingroup$ In proposition 1 you have a divergent series! $\endgroup$ – Zaid Alyafeai Feb 28 '17 at 14:24
  • $\begingroup$ We do have the generating function for the Hurwitz zeta, so it's safe to integrate from 0 to x. And if you are talking about $\dfrac {1}{n+a+x} $ then wikipidea says it's a way to write Digamma. $\endgroup$ – Aditya Narayan Sharma Feb 28 '17 at 14:51

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