18
$\begingroup$

While studying Ramanujan's Collected Papers I came across a paper titled "Some Definite Integrals" which appeared in Messenger of Mathematics, XLIV, 1915, 10-18. It contains lot of weird integrals for which Ramanujan has given proofs. However in one instance he discusses about the integral $$\int_{0}^{\infty}\frac{dx}{(1 + x^{2})(1 + r^{2}x^{2})(1 + r^{4}x^{2})\cdots} = \frac{\pi}{2(1 + r + r^{3} + r^{6} + r^{10} + \cdots)}\tag{1}$$ where $0 < r < 1$. Ramanujan derives this formula from $$\int_{0}^{\infty}\frac{(1 + arx)(1 + ar^{2}x)\cdots}{(1 + x)(1 + rx)(1 + r^{2}x)\cdots}x^{n - 1}\,dx = \frac{\pi}{\sin n\pi}\prod_{m = 1}^{\infty}\frac{(1 - r^{m - n})(1 - ar^{m})}{(1 - r^{m})(1 - ar^{m - n})}\tag{2}$$ where $0 < r < 1, n > 0, 0 < a < r^{n - 1}$ and $n$ is not an integer and $a$ is not of the form $a = r^{p}$ where $p$ is a positive integer. Unfortunately Ramanujan does not prove the formula $(2)$.

Is there any direct approach to establish $(1)$ without using $(2)$ or some way to establish $(2)$?

$\endgroup$
  • $\begingroup$ Isn't it possible by Complex Analysis? Indeed, maybe it's good to integrate $f(z)$ over a semi circle in upper half plane and then apply Residue Theorem to gain the formula and at last, take real parts and send radius to infinity. $\endgroup$ – Fardad Pouran May 25 '14 at 10:21
  • $\begingroup$ @fardad.math: I am not well versed in complex analysis (as you may observe from my questions and answers on this site). Is there is any approach within the framework of real variables? $\endgroup$ – Paramanand Singh May 25 '14 at 16:32
10
$\begingroup$

You can use residues, let $f(z)=\frac{1}{(1+z^2)(1+r^2z^2)(1+r^4z^2)...}$ this has an infite set of singularities at $z=\pm i 1/r^{n},n\in\Bbb N_0$. We can see that $\large\int_0^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx=\frac{1}{2}\int_{-\infty}^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx$

So we will consider the upper semi circle contour, spanning over $-\infty$ to $+\infty$, only the positive singularities are in this contour, i.e. $z=i\frac{1}{r}$.

Now $\operatorname{Res}(f(z),z=\frac{i}{r^{n}})=\large\lim_{z\to i\frac{1}{r^{n}}}(z-i\frac{1}{r^{n}})f(z)=\large\lim_{z\to i\frac{1}{r^{n}}}\frac{1}{r^n}(r^nz-i)f(z)$

$=\lim_{z\to i\frac{1}{r^{n}}}\frac{1}{r^n(r^nz+i)}\prod_{j=0,j\ne n}^\infty\frac{1}{(1+x^{2}r^{2j})}$

$=\large\frac{1}{2ir^n}\prod_{j=0,j\ne n}^\infty\frac{1}{(1-r^{2j-2n})}$

Now $\frac{1}{2}\int_{-\infty}^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx=\frac{1}{2}2\pi i\sum \operatorname{Res}(f)=\frac{1}{2}\pi i\sum_{n=0}^\infty\frac{1}{ir^n}\prod_{j=0,j\ne n}^\infty\frac{1}{(1-r^{2j-2n})}$

From here I'm not sure how to reach the closed form, but hopefully this helps to show a different approach, even if you are not too well versed in integrals by residue :)

$\endgroup$
  • $\begingroup$ I believe there is some typo. The function has terms of the form $(1 + r^{2n}x^{2})$ and not $(1 + r^{2n}x^{2n})$. And then the singularities are of the form $\pm i/r^{n}$. $\endgroup$ – Paramanand Singh May 26 '14 at 7:38
  • $\begingroup$ @ParamanandSingh, thank you! I have edited it, hopefully all is good now :) $\endgroup$ – Ellya May 26 '14 at 7:50
  • $\begingroup$ I will try to simplify the last expression you obtained and try to get the expression by Ramanujan. $\endgroup$ – Paramanand Singh May 26 '14 at 7:53
  • $\begingroup$ @ParamanandSingh, hopefully it will work out nicely :) $\endgroup$ – Ellya May 26 '14 at 8:07
  • $\begingroup$ @ParamanandSingh, I noticed some other mistakes, which I have corrected :) $\endgroup$ – Ellya May 26 '14 at 9:03
4
$\begingroup$

I find this relationship charming.

Here is what I have so far as an alternative solution without resorting to complex analysis. For $n=0,1,...$ define \begin{align} I_n=\int_0^{\infty} \frac{dx}{(1+x^2)(1+r^2x^2)...(1+r^{2n}x^2)} \end{align} Doing many partial fractions I could establish for $n=1,2,3,4,...$ that \begin{align} I_{n}&=\frac{1-r^{2n-1}}{1-r^{2n}}I_{n-1}\\ I_0&=\frac{\pi}{2} \end{align} Therefore \begin{align} I_{\infty}&=\frac{\prod_{n=1}^{\infty}\Big(1-r^{2n-1}\Big)}{\prod_{n=1}^{\infty}\Big(1-r^{2n}\Big)}\frac{\pi}{2} \end{align} Now using the same arguments as in this post, we can establish $(1)$.

$\endgroup$
  • $\begingroup$ I like your totally elementary answer. +1 $\endgroup$ – Paramanand Singh Feb 19 '15 at 8:50
  • $\begingroup$ @ParamanandSingh Many thanks for the comment. Do you know where could I read more on the works of Ramanujan? Are his notes published and accessible? Thanks. $\endgroup$ – Math-fun Feb 19 '15 at 9:51
  • $\begingroup$ You should try to get access to the following : Ramanujan Notebooks Vol 1-5 by Bruce C. Berndt, Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work by G. H. Hardy and the best work is "Collected Papers of Ramanujan". The book "Pi and the AGM" by Borwein Brothers is also very helpful. I have tried to present some of the Ramanujan stuff in a systematic fashion in a series of posts in my blog. You can start with paramanands.blogspot.com/2011/11/… and read further posts after that. $\endgroup$ – Paramanand Singh Feb 19 '15 at 9:58
  • $\begingroup$ @ParamanandSingh Many thanks for the helpful hints. I will start by taking a look at your works. $\endgroup$ – Math-fun Feb 19 '15 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.