Proof of an integral identity involving $\pi$ and e

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 $s$!!