<s>[INST] Show that $\int_0^\infty\frac{1}{1+x^2+x^{\pi}+x^{\pi-2}}dx = \int_0^\infty\frac{1}{1+x^2+x^{e}+x^{e-2}}dx $ Vladimir Reshetnikov's Identity
(1)
$$\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx  =\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{e}}dx  $$
Note that $(1+x^2)(1+x^{\pi})=1+x^2+x^{\pi}+x^{2+\pi}$ and
$(1+x^2)(1+x^e)=1+x^2+x^e+x^{2+e}$
**An imitation of Vladimir Reshetnikov's identity **
(2)
$$\int_0^\infty\frac{1}{1+x^2+x^{\pi}+x^{\pi-2}}dx = \int_0^\infty\frac{1}{1+x^2+x^{e}+x^{e-2}}dx $$
The problem with this denominator $(1++x^2+x^{\pi}+x^{\pi-2})$ we can't factorised it
The proved of Identity (1) was offered by Dr.MV
I was hoping to follow the method of Dr.MV of prove of (1) and use it in (2) but its denominator wasn't factorisable, so I am stuck at this point. Can anybody offer their own way of proving identity (2)

My try:
Let (2) be writen in the general form $$I=\int_0^\infty\frac{1}{1+x^2+x^a+x^{a-2}}dx $$
I change (2) into $$I=\int_0^\infty\frac{1}{(1+x^{-2})(1+x^a)+x^2-x^{-2}}dx$$ from here I still think it is more harder then before, anyway I am stuck. **I need help **
 [/INST] Let we do something more general. Let $a>0$ and 
$$ I(a) = \int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^a)}. $$
Well, $I(a)$ is constant. Don't you believe it? Replace $x$ with $\frac{1}{z}$ to get:
$$ I(a) = \int_{0}^{+\infty}\frac{z^a\,dx}{(1+z^2)(1+z^a)}\,dz $$
from which:
$$ I(a) = \frac{1}{2}\int_{0}^{+\infty}\frac{1+x^a}{(1+x^2)(1+x^a)}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{dx}{1+x^2}=\color{red}{\frac{\pi}{4}}.$$
You integrals are just $I(e-2)$ and $I(\pi-2)$, and they both equal $\frac{\pi}{4}$.
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