11
$\begingroup$

In The Man Who Knew Infinity, Kanigel discusses Ramanujan's original letter to Hardy. In it, Ramanujan states that if $\alpha\beta=\pi^2$, then $$ \alpha^{-1/4}\left(1+4\alpha\int_0^\infty\frac{xe^{-\alpha x^2}}{e^{2\pi x}-1} \, dx \right)=\beta^{-1/4}\left(1+4\beta\int_0^\infty\frac{xe^{-\beta x^2}}{e^{2\pi x}-1} \, dx\right). $$ Does anyone know of a reference to a proof of this fact? There is no evidence offered in the book, other than a "Hardy had proved theorems like it, had even offered a similar one as a mathematical question...years before."

I am pretty unfamiliar with Ramanujan's work, so I don't really know where a good starting point would be for such a problem. Possibly using contour integration to turn it into an infinite sum via residues? I am a bit skeptical of this, as I have heard that Ramanujan was not very familiar with complex analysis at all.

I'm also not sure what the best way to search for this would be, despite looking for "Ramanujan integral identities," of which there are many to search through...

Thanks in advance for any help.

$\endgroup$
3
  • $\begingroup$ I can't help with the reference, but is it possible that you have a typo on one side or the other? Those expressions look to me like they are identical, just alpha replaced with beta. $\endgroup$ Jun 15, 2017 at 1:30
  • $\begingroup$ @JasonKnapp That is correct (at least as written in the book); that is in part why I am so surprised $\endgroup$
    – TomGrubb
    Jun 15, 2017 at 1:31
  • 1
    $\begingroup$ The similar theorems of Hardy that Kanigel refers to are probably from this paper. It's worth checking the notes on the chapter in TMWKI: Kanigel's good at citing his sources properly. $\endgroup$
    – Chappers
    Jun 15, 2017 at 2:29

2 Answers 2

7
$\begingroup$

Start with the Bernoulli generating function $$\sum_{n=0}^{\infty} \frac{B_{n} \, x^n}{n!} = \frac{x}{e^{x} - 1}$$ to obtain \begin{align} I &= \int_{0}^{\infty} \frac{x \, e^{-\alpha \, x^{2}}}{e^{2\pi x} - 1} \, dx \\ 4 \pi^{2} \, I &= \int_{0}^{\infty} \frac{u \, e^{- p \, u^2}}{e^{u} - 1} \, dx \hspace{5mm} \text{ $u = 2\pi x$ and $p = \frac{\alpha}{4 \pi^{2}}$} \\ &= \sum_{n=0}^{\infty} \frac{B_{n}}{n!} \, \int_{0}^{\infty} e^{- p \, u^2} \, u^{n} \, dx \\ &= \frac{1}{2 \sqrt{p}} \, \sum_{n=0}^{\infty} \frac{B_{n}}{n!} \, \left(\frac{1}{p}\right)^{n/2} \, \Gamma\left(\frac{n+1}{2}\right) \\ &= \frac{1}{2 \sqrt{p}} \, \left[ - \frac{1}{2 \sqrt{p}} + \sqrt{\pi} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{1}{4 p}\right)^{n} \right] \\ &= - \frac{\pi^{2}}{\alpha} + \pi \, \sqrt{\frac{\pi}{\alpha}} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{\pi^{2}}{\alpha}\right)^{n} \end{align}

This leads to \begin{align} \alpha^{-1/4}\left(1+4\alpha\int_0^\infty\frac{xe^{-\alpha x^2}}{e^{2\pi x}-1}dx\right) = \frac{\alpha^{1/4}}{\sqrt{\pi}} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{\pi^{2}}{\alpha}\right)^{n}. \end{align}

From here it is just making the connection.

$\endgroup$
1
  • 3
    $\begingroup$ The Bernoulli series only converges for $|x|<2\pi$, so how can you replace the generating function with the series and integrate from zero to infinity? $\endgroup$
    – Dave
    Jul 9, 2021 at 11:42
5
$\begingroup$

I believe Ramanujan used self-Fourier transforms to prove this, as he did for another similar formula in his first letter. Consider the two sine transforms $$ \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \sin(2\pi xt) \, dt = \frac{1}{2} \left(\frac{1}{e^{2\pi x}-1} - \frac{1}{2\pi x} \right), \hspace{0.5cm} x>0 $$ and $$ \int_0^{\infty} te^{-\pi \alpha t^2} \sin(2\pi xt) \, dt = \frac{1}{2} \alpha^{-3/2} xe^{-\pi x^2/\alpha}, \hspace{0.5cm} x,\; \text{Re }\alpha >0. $$ Applying the Fubini theorem we have $$ \int_0^{\infty} \left(\frac{1}{e^{2\pi x}-1}-\frac{1}{2\pi x} \right) xe^{-\pi \alpha x^2} \, dx = \int_0^{\infty} 2x e^{-\pi \alpha x^2} \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \sin(2\pi xt) \, dt \, dx $$ $$ = \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \int_0^{\infty} 2x e^{-\pi \alpha x^2} \sin(2\pi xt) \, dx \, dt = \alpha^{-3/2} \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) t e^{-\pi t^2/\alpha} \, dt $$ and therefore $$ \int_0^{\infty} \frac{xe^{-\pi \alpha x^2}}{e^{2\pi x}-1} \, dx - \frac{1}{4\pi \alpha^{1/2}} = \alpha^{-3/2} \int_0^{\infty} \frac{te^{-\pi t^2/\alpha}}{e^{2\pi t}-1} \, dt - \frac{1}{4\pi \alpha}, \hspace{0.5cm} \text{Re } \alpha>0 $$ where we have evaluated the two Gaussian integrals. Replacing $\pi \alpha$ with $\alpha$ and $\pi/\alpha$ with $\beta$ gives Ramanujan's symmetric expression.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .