<s>[INST] Real-Analysis Methods to Evaluate $\int_0^\infty \frac{x^a}{1+x^2}\,dx$, $|a|<1$. 
In THIS ANSWER, I used straightforward contour integration to evaluate the integral $$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^a}{1+x^2}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)}$$for $|a|<1$.

An alternative approach is to enforce the substitution $x\to e^x$ to obtain
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\
&=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\
&=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\
&=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\
&=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \tag 1\\\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2
\end{align}$$
Other possible ways forward include writing the integral of interest as
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{0}^1 \frac{x^{a}+x^{-a}}{1+x^2}\,dx
\end{align}$$
and proceeding similarly, using $\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$.

Without appealing to complex analysis, what are other approaches one can use to evaluate this very standard integral?

EDIT:

Note that we can show that $(1)$ is the partial fraction representation of $(2)$ using Fourier series analysis.  I've included this development for completeness in the appendix of the solution I posted on THIS PAGE.

 [/INST] This is an elaboration of the last step in Dr. MV's original solution.
The Mittag-Leffler expansion of the secant function is
$$\sec(z) = \pi \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(\pi /2)^{2}(2n+1)^{2} - z^{2}}$$ 
Thus we have
\begin{align}
\frac{\pi}{2} \sec\left(\frac{\pi a}{2} \right) 
&= \frac{\pi ^{2}}{2} \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(\pi /2)^{2}(2n+1)^{2} - (\pi a/2)^{2}} \\
&= 2 \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(2n+1)^{2} - a^{2}}
\end{align}
</s>