The following identity is allegedly due to Ramanujan $$\int_0^\infty \frac{{\rm d}x}{(1+x^2)(1+r^2x^2)(1+r^4x^2)\cdots} = \frac{\pi/2}{\sum_{n=0}^\infty r^{\frac{n(n+1)}{2}}} \, $$but how do you prove this? The denominator of the right side is related to the Jacobi Function, so maybe one could proceed via modular forms?
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3The residue theorem and Jacobi triple product should lead to a simple answer. – Jack D'Aurizio Aug 17 '20 at 21:05
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Well I actually tried that, but somehow it didn't really help. What I had was $$\prod_{n=1}^\infty (1-r^{2n})(1+r^n) = \sum_{n=0}^\infty r^{n(n+1)/2} $$ by setting $x=r^{1/2}$ and $y=r^{1/4}$. – Diger Aug 17 '20 at 21:37
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Using the residue theorem (after extending to $-\infty$) and closing above one obtains $$\frac{1}{2} \int_{-\infty}^\infty \frac{{\rm d}x}{\prod_{n=0}^\infty \left[1+(xr^n)^2\right]} = \pi i \sum_{k=0}^\infty {\rm Res} \frac{1}{\prod_{n=0}^\infty \left[1+(xr^n)^2\right]} \Bigg|{x=\frac{i}{r^k}} = \frac{\pi}{2} \sum{k=0}^\infty \frac{1}{r^k} \prod_{\substack{n=0 \ n\neq k}}^\infty \frac{1}{1-r^{2n-2k}} , .$$ – Diger Aug 18 '20 at 08:12
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$$=\frac{\pi}{2} \sum_{k=0}^\infty \frac{1}{r^k} \left(\prod_{n=1}^k \frac{1}{1-r^{-2n}}\right) \left(\prod_{n=1}^\infty \frac{1}{1-r^{2n}}\right) = \frac{\pi}{2} \frac{\prod_{n=1}^\infty (1+r^n)}{\sum_{n=0}^\infty r^{n(n+1)/2}} \sum_{k=0}^\infty \frac{1}{r^k} \prod_{n=1}^k \frac{1}{1-r^{-2n}} , .$$ But how do I show that the product in the nominator and right sum cancel? – Diger Aug 18 '20 at 08:55
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So essentially we have to prove the identity $$\prod_{n=1}^\infty \frac{1}{1+r^n} = \sum_{k=0}^\infty \frac{1}{r^k} \prod_{n=1}^k \frac{1}{1-r^{-2n}} , .$$ – Diger Aug 18 '20 at 09:08
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Do you mean for the latter identity or for the original problem? – Diger Aug 18 '20 at 10:31
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When expanding out the LHS I'll get $$\sum_{m=0}^\infty r^m \sum_{\substack{n_1,n_2,n_3,\cdots=0 \ n_1+2n_2+3n_3+\cdots=m}}^\infty (-1)^{n_1+n_2+\cdots}$$ and similarly expanding out the RHS gives $$\sum_{m=0}^\infty r^m \substack{\sum_{k=0}^\infty (-1)^k \sum_{n_1,\cdots,n_k=0}^\infty 1 \ k^2+2(n_1+2n_2+\cdots+kn_k)=m}$$ but I don't see a "trivial" way why the counting for the coefficients should be equal. – Diger Aug 18 '20 at 11:22
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A partial answer for now. We have to prove that $$ \prod_{n\geq 1}\frac{1}{1+r^n}=\sum_{k\geq 0}\prod_{n=1}^{k}\frac{r^{2n-1}}{r^{2n}-1} $$ or $$ \prod_{n\geq 1}\frac{1-r^n}{1-r^{2n}}=\sum_{k\geq 0}(-1)^k r^{k^2} \prod_{n=1}^{k}\frac{1}{1-r^{2n}} $$ or $$ \prod_{n\geq 1}(1-r^n) = \sum_{k\geq 0}(-1)^k r^{k^2} \prod_{n>k}(1-r^{2n}) $$
where the LHS, by Euler's pentagonal number theorem, equals $$\sum_{k=-\infty}^{+\infty}(-1)^k r^{k(3k-1)/2} $$ and the coefficient of $r^m$ in $\prod_{n>k}(1-r^n)$ depends on the number of partitions of $m$ into distinct parts with cardinality $>k$, accounted with a positive or negative sign according to the number of parts.
Now it shouldn't be difficult to prove our claim by using the same involution exploited in the combinatorial proof of Euler's pentagonal number theorem, or something quite close to it.
Jack D'Aurizio
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Hey, thanks for your effort. Just some minor things: In the first formula it should be $r^{2n-1}$ instead of $r^{2n-k}$ if you pull it inside the product. In the second formula therefore it should be $r^{k^2}$ instead of $r^k$. I'll think about the rest. – Diger Aug 18 '20 at 17:58
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Thank you. Interesting how the proper solution by math-fun is actually not the one accepted. – Diger Aug 25 '20 at 17:43