<s>[INST] What is the closed form for $\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{1}{1+x^ e}dx $? On my previou page Jack D'Aurizio offered a concise elegant prove of Vladimir Reshetnikov's identity and a closed form for it.
(1)
$$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx=\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{1}{1+x^{e}}dx=\frac{\pi}{4}$$
Here we have another imitation of Vladimir Reshetnikov's identity
(2)
$$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{1}{1+x^e}dx =\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^e}{1+x^e}\cdot\frac{1}{1+x^{\pi}}dx$$
A closed form of (2) is unknown
We ask if this identity (2) can be proven in the same way as (1) and with a closed form.
 [/INST] Through the substitution $x\mapsto\frac{1}{x}$ we have:
$$ I(a,b) = \int_{0}^{+\infty}\frac{x^a}{1+x^a}\cdot\frac{1}{1+x^b}\cdot\frac{dx}{1+x^2} = \int_{0}^{+\infty}\frac{1}{1+x^a}\cdot\frac{x^b}{1+x^b}\cdot\frac{dx}{1+x^2} = I(b,a) $$
hence, by your previous question:
$$\begin{eqnarray*} I(a,b) &=& \frac{1}{2}\int_{0}^{+\infty}\frac{x^a+x^b}{(1+x^a)(1+x^b)}\cdot\frac{dx}{1+x^2}\\&=&\frac{1}{2}\left[\int_{0}^{+\infty}\left(\frac{1}{1+x^a}+\frac{1}{1+x^b}\right)\frac{dx}{1+x^2}-2\int_{0}^{+\infty}\frac{dx}{(1+x^a)(1+x^b)(1+x^2)}\right]\\&=&\frac{\pi}{4}-\int_{0}^{+\infty}\frac{dx}{(1+x^a)(1+x^b)(1+x^2)}\end{eqnarray*}$$
that, at least in principle, can be computed from partial fraction decomposition, the residue theorem and the identity:
$$ \int_{0}^{+\infty}\frac{x^b dx}{1+x^a} = \frac{\pi}{a\sin\left(\frac{\pi (b+1)}{a}\right)}.$$ 
</s>