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.

 A: Hint. Assume $|a|<1$. Another equivalent approach would be to write
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_0^\infty x^a \left(\int_0^\infty e^{-(1+x^2)t}dt\right)dx
\\\\&=\int_0^\infty e^{-t}\left(\int_0^\infty x^a e^{-tx^2}dx\right)dt\\\\
&=\frac12\Gamma\left(\frac{1+a}{2}\right)\int_0^\infty t^{\frac{1-a}{2}-1}e^{-t}dt\\\\
&=\frac12\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(\frac{1-a}{2}\right)\\\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)
\end{align}$$ by using the standard integral representation of the $\Gamma$ function and (6.1.30).
A: $$
\begin{align}
\int_0^\infty\frac{x^a}{1+x^2}\,\mathrm{d}x
&=\frac12\int_0^\infty\frac{x^{(a-1)/2}}{1+x}\,\mathrm{d}x\\
&=\frac12\int_0^1\frac{\left(\frac{t}{1-t}\right)^{(a-1)/2}}{1+\left(\frac{t}{1-t}\right)}\frac{\mathrm{d}t}{(1-t)^2}\\
&=\frac12\int_0^1t^{(a-1)/2}(1-t)^{(-1-a)/2}\,\mathrm{d}t\\
&=\frac12\mathrm{B}\left(\frac{1+a}2,\frac{1-a}2\right)\\
&=\frac12\frac{\Gamma\left(\frac{1+a}2\right)\Gamma\left(\frac{1-a}2\right)}{\Gamma(1)}\\
&=\frac\pi2\sec\left(\frac{\pi a}2\right)
\end{align}
$$
As in kobe's answer, we've used Euler's Reflection Formula. However, most proofs of that I've seen use contour integration.
A: I'll assume $\lvert a\rvert < 1$. Letting $x = \tan \theta$, we have
$$\int_0^\infty \frac{x^a}{1 + x^2}\, dx = \int_0^{\pi/2}\tan^a\theta\, d\theta = \int_0^{\pi/2} \sin^a\theta \cos^{-a}\theta\, d\theta$$
The last integral is half the beta integral $B((a + 1)/2, (1 - a)/2)$, Thus
$$\int_0^{\pi/2}\sin^a\theta\, \cos^{-a}\theta\, d\theta = \frac{1}{2}\frac{\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{a+1}{2} + \frac{1-a}{2}\right)} = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)$$
By Euler reflection,
$$\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \pi \csc\left[\pi\left(\frac{1+a}{2}\right)\right] = \pi \sec\left(\frac{\pi a}{2}\right)$$
and the result follows.
Edit: For a proof of Euler reflection without contour integration, start with the integral function $f(x) = \int_0^\infty u^{x-1}(1 + u)^{-1}\, du$, and show that $f$ solves the differential equation $y''y - (y')^2 = y^4$, $y(1/2) = \pi$, $y'(1/2) = 0$. The solution is $\pi \csc \pi x$. On the other hand, $f(x)$ is the beta integral $B(1+x,1-x)$, which is equal to $\Gamma(x)\Gamma(1-x)$. I believe this method is due to Dedekind.
A: \begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&= \int_0^\infty x^a  \int_0^\infty e^{-xt} \sin  t \,dt\, dx\\
&=\int_0^\infty  \sin  t \int_0^\infty e^{-xt} x^a \,dx\, dt\\
&=\Gamma (a+1)\int_0^\infty t^{-a-1}  \sin t \,dt\\
&=-\Gamma (a+1)\Gamma (-a) \sin (\pi a/2)\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)
\end{align}
A: 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}
