9
$\begingroup$

While browsing through some letters of Ramanujan to G. H. Hardy I came across these two formulas (which are as mysterious as any other formula of Ramanujan): $$4\int_{0}^{\infty}\frac{xe^{-x\sqrt{5}}}{\cosh x}\,dx = \cfrac{1}{1+}\cfrac{1^{2}}{1+}\cfrac{1^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{3^{2}}{1+}\cfrac{3^{2}}{1+}\cdots\tag{1}$$ $$2\int_{0}^{\infty}\frac{x^{2}e^{-x\sqrt{3}}}{\sinh x}\,dx = \cfrac{1}{1+}\cfrac{1^{3}}{1+}\cfrac{1^{3}}{3+}\cfrac{2^{3}}{1+}\cfrac{2^{3}}{5+}\cfrac{3^{3}}{1+}\cfrac{3^{3}}{7+}\cdots\tag{2}$$ From these it appears that Ramanujan had general formulas for the integrals $$I_{m, n} = \int_{0}^{\infty}\frac{x^{m}e^{-x\sqrt{n}}}{\cosh x}\,dx,\,J_{m, n} = \int_{0}^{\infty}\frac{x^{m}e^{-x\sqrt{n}}}{\sinh x}\,dx\tag{3}$$ and for some values of $m, n$ he was able to express these as continued fractions.

I am totally perplexed by these formulas. Do we have any proofs of these or perhaps any reference which treats the integrals $I_{m, n}, J_{m, n}$?


Just for clarity the notation $$\cfrac{a_{1}}{b_{1}+}\cfrac{a_{2}}{b_{2}+}\cfrac{a_{3}}{b_{3}+}\cdots$$ consumes less space in typing compared to the more cumbersome but easier to understand continued fraction $$\cfrac{a_{1}}{b_{1} + \cfrac{a_{2}}{b_{2} + \cfrac{a_{3}}{b_{3} + \cdots}}}$$

$\endgroup$
  • $\begingroup$ According to Berndt, (1) was proved by Stieltjes and later by Rogers (Ramanujan's Notebooks, II, page 151). (2) is due to Stieltjes (Ramanujan's Notebooks, II, page 153). Though there are parametric extensions of (1) and (2) I doubt that generalizations similar to (3) exist. $\endgroup$ – user82588 Jan 17 '17 at 18:56
  • $\begingroup$ @Nemo: Thanks. I will have a look at the references you gave. $\endgroup$ – Paramanand Singh Jan 17 '17 at 21:04
  • 4
    $\begingroup$ @Algebra: it is a bit strange that you are totally unaware of the great Indian mathematician Ramanujan who was as famous as an Indian could be. His works are comparable in quality to people like Euler and Jacobi. See en.wikipedia.org/wiki/Srinivasa_Ramanujan $\endgroup$ – Paramanand Singh Jan 18 '17 at 7:25
  • 1
    $\begingroup$ Tragic death though IMO. :'( There was more awaiting the great one... he died too young. $\endgroup$ – Simply Beautiful Art Jan 20 '17 at 1:11
  • 2
    $\begingroup$ @SimpleArt : fully agree. Had he lived longer, perhaps we could expect some marvelous proofs. Most of the proofs of difficult theorems of Ramanujan (as given by Berndt and other authors) are mere verification using sophisticated tools of complex analysis (read modular forms). But the few proofs Ramanujan offered were very economical and used basic algebra and calculus. I have scribbled my thoughts on Ramanujan in this post paramanands.blogspot.com/2013/06/thoughts-on-ramanujan.html $\endgroup$ – Paramanand Singh Jan 20 '17 at 5:09
2
$\begingroup$

The following continued fraction is given on page 151 of Berndt's book "Ramanujan's Notebooks", Vol. II:

enter image description here

Let $x=\sqrt{5}$, and $n\to 0$ to obtain \begin{align} \int_{0}^{\infty}e^{-u\sqrt{5}}\frac{u}{\cosh u}\,du &= \cfrac{1}{4+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{4+}\cfrac{4^{2}}{1+}\cfrac{4^{2}}{4+}\cfrac{6^{2}}{1+}\cfrac{6^{2}}{4+}\ldots\\ &=\frac14 \cdot \cfrac{1}{1+}\cfrac{1^{2}}{1+}\cfrac{1^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{3^{2}}{1+}\cfrac{3^{2}}{1+}\ldots \end{align}


$(2)$ is obtained from the continued fraction (32.4) on page 153 enter image description here$\tag{32.4}$

Note that \begin{align} 2\int_{0}^{\infty}\frac{u^{2}e^{-u\sqrt{3}}}{\sinh u}\,du&=\sum_{n=0}^\infty 4\int_{0}^{\infty}u^{2}e^{-u\sqrt{3}-u-2nu}\,du\\ &=8\sum_{n=0}^\infty\frac{1}{(2n+1+\sqrt3)^3}\\ &=\sum_{n=1}^\infty\frac{1}{\left(n+\tfrac{\sqrt3-1}{2}\right)^3}. \end{align}

Therefore (32.4) with $x=\frac{\sqrt3-1}{2}$ (i.e. $2x(x+1)=1$) gives $$ 2\int_{0}^{\infty}\frac{x^{2}e^{-x\sqrt{3}}}{\sinh x}\,dx = \cfrac{1}{1+}\cfrac{1^{3}}{1+}\cfrac{1^{3}}{3+}\cfrac{2^{3}}{1+}\cfrac{2^{3}}{5+}\cfrac{3^{3}}{1+}\cfrac{3^{3}}{7+}\cdots $$

| cite | improve this answer | |
$\endgroup$

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.