TITLE: How find this integral $\int_{0}^{\infty}\frac{dx}{(1+x^2)(1+r^2x^2)(1+r^4x^2)(1+r^6x^2)\cdots}$
QUESTION [16 upvotes]: 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 $F(y)=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)(1+(x+y)^2)}$

REPLY [14 votes]: 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:
$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$