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prove that this integral

$$\int_{0}^{\infty}\dfrac{dx}{(1+x^2)(1+r^2x^2)(1+r^4x^2)(1+r^6x^2)\cdots}= \dfrac{\pi}{2(1+r+r^3+r^6+r^{10}+\cdots}$$

for this integral,I can't find it.and I don't know how deal this such strange integral.

and this problem is from china QQ (someone ask it)

before I ask this question: How find this integral $F(y)=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)(1+(x+y)^2)}$

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    $\begingroup$ Induction seems like a sound if unimaginative bet: find a recursion relation which gives the integral over $n+1$ such products in terms of that of $n$ products, then solve the recurrence and take $n\to\infty$. The residue theorem could also provide an approach to such a formula. $\endgroup$ – Semiclassical Jul 23 '14 at 17:29
  • $\begingroup$ Have you tried thinking of the LHS as f(r), taking logs / using the Leibniz rule? $\endgroup$ – user_of_math Jul 23 '14 at 17:36
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    $\begingroup$ Also, the series on the RHS is a rather interesting function: $\sum_{k=0}^\infty r^{n(n+1)/2}=\vartheta_2(0,r^{1/2})/(2r^{1/8})$ where $\vartheta_2(0,x)$ is a Jacobi theta function. (WolframAlpha) $\endgroup$ – Semiclassical Jul 23 '14 at 17:48
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    $\begingroup$ This exercise reminded me more the Jacobi Triple Product and the Euler identities. I believe that the Problem 3 in sporadic.stanford.edu/bump/Math108/jtp.pdf and the residue theorem lead to a proof. $\endgroup$ – Jack D'Aurizio Jul 23 '14 at 20:27
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    $\begingroup$ It worked and it is a quite interesting trick: we first convert a sum (the sum of residues) into a product through the Euler's identity, then convert back the product into the reciprocal of a sum through the Jacobi triple product identity. $\endgroup$ – Jack D'Aurizio Jul 23 '14 at 21:32
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If we set $$ f(x)=\prod_{n=0}^{+\infty}(1+r^{2n}x^2) $$ we have: $$\int_{0}^{+\infty}\frac{dx}{f(x)}=\pi i\sum_{m=0}^{+\infty}\operatorname{Res}\left(f(z),z=\frac{i}{r^m}\right)=\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}\tag{1}$$ but since $$ \prod_{n=0}^{+\infty}(1-x^n z)^{-1}=\sum_{n=0}^{+\infty}\frac{z^n}{(1-x)\cdot\ldots\cdot(1-x^n)}$$ is one of the Euler's partitions identities, and: $$\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\sum_{m=0}^{+\infty}\frac{(1/r)^m}{(1-(1/r^2))\cdot\ldots\cdot(1-(1/r^2)^m)}$$ we have: $$\int_{0}^{+\infty}\frac{dz}{f(z)}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\prod_{m=0}^{+\infty}\left(1-\frac{1}{r^{2m+1}}\right)^{-1}\tag{2}$$ and the claim follows from the Jacobi triple product identity: $$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$

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    $\begingroup$ Marvelous. Now I just wish I knew I generating-function interpretation of the identity, but that's probably too much too hope for. $\endgroup$ – Semiclassical Jul 23 '14 at 21:43
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    $\begingroup$ @Semiclassical: To remember the details, I just googled and found the notes of Professor D.M.Jackson - www.math.uwaterloo.ca/~dmjackso/CO630/JTPID.pdf - they are written pretty good. $\endgroup$ – Jack D'Aurizio Jul 23 '14 at 21:48

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