Proof of an integral identity involving $\pi$ and e In the "Surprising Identities" post from a while back, Vladimir Reshetnikov offered the following identity[1]:
$$\int_{0}^{\infty} dx \frac{1}{1 + x^2} \frac{1}{1 + x^{\pi}} = \int_{0}^{\infty}dx  \frac{1}{1 + x^2} \frac{1}{1 + x^{e}} \ \ .$$
Where can I find a proof of this?


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*https://math.stackexchange.com/a/693459/29360
 A: Observe the integral
$$\int_0^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx$$for any exponent $s$ for which the integral converges.  Then, split the integral $I$ as 
$$I= \int_0^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_1^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx$$
In the second integral, make the substitution $x=\frac{1}{y}$, $dx=-\frac{1}{y^2}dy$, and note that the limits of integration transform from $(1,\infty)$ to $(1,0)$.  Thus, we can write
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_1^{0} \frac{1}{1+y^{-2}}\frac{1}{1+y^{-s}} \left(-\frac{1}{y^2}\right) dy$$
In the last integral, absorbing the negative sign by interchanging the order of integration, multiplying numerator and denominator by $y^2y^s$ and simplifying, and changing the dummy integration variable to x yields
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_0^{1} \frac{1}{1+x^{2}}\frac{x^s}{1+x^{s}} dx$$
which after recombining the integrals reveals that
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1+x^s}{1+x^{s}}dx=\int_0^{1} \frac{1}{1+x^2}dx=\frac{\pi}{4}$$which obviously is independent of $s$!!
A: $$\int_0^{+\infty}\frac{dx}{(1+x^2)(1+x^s)}\stackrel{x=\tan{\theta}}{=}\int_0^{\pi/2}\frac{d\theta}{1+\tan^s\theta}\\ \qquad = \int_0^{\pi/2}\frac{\sin^s\theta}{\cos^s \theta+\sin^2\theta}\ d\theta\\ \qquad\stackrel{\phi=\pi/2-\theta}{=}\int_0^{\pi/2}\frac{\cos^s \phi}{\sin^s\phi +\cos^s \phi}\ d\phi\\ =\frac{\pi}{4}$$
where the last line follows by summing the two previous lines together and dividing by $2$.
