Symmetry of function defined by integral Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as
$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
One can use, for example, the Residue Theorem to show that
$$ f(\alpha, \beta) = \frac{\pi \sin{\left (\pi \alpha \beta\right )}}{ \sin{\left (\pi \alpha\right )} \,  \sin{\left (\pi \beta\right )}} $$
Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$.  My question is, can one see this symmetry directly from the integral expression?

Very interesting question! But, alas, not an answer. Only few representations for the integral obtained. One of them evaluated to the form claimed in the question.

First, transform the integral into a form, symmetric under $\alpha \mapsto -\alpha$:
$$
  \int_0^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x = \int_0^1 \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x  + \int_1^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x 
$$
Make a change of variables $x \to x^{-1}$ in the last integral to obtain:
$$
   f(\alpha,\beta) = \int_0^1  \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1}
$$
Now, making a change of variables $x = \exp(-t)$ we have:
$$
  f(\alpha,\beta) =  \int_0^\infty \frac{\cosh(\alpha t)}{\cosh(t) + \cos(\beta \pi)} \mathrm{d} t  \tag{2}
$$
Using 
$$
   \int_0^\infty \exp\left(-u \left( \cosh t + \cos \pi \beta \right) \right) \mathrm{d}u = \frac{1}{\cosh(t) + \cos(\beta \pi)}
$$
and the integral representation of the modified Bessel function of the second kind:
$$
    \int_0^\infty \cosh(\alpha t) \exp\left( - u \cosh t \right) \mathrm{d}t = K_\alpha(u)
$$
we arrive at a compact representation:
$$
  f(\alpha,\beta) = \int_0^\infty K_\alpha(u) \mathrm{e}^{-u \cos\left(\pi \beta\right)} \mathrm{d} u \tag{3}
$$
expanding the exponential into series and using $\int_0^\infty u^n K_\alpha(u) \mathrm{d} u = 2^{n-1} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right)$ we get:
$$
   f(\alpha,\beta) = \sum_{n=0}^\infty \frac{2^{n-1}}{n!} \left(-\cos \pi \beta\right)^{n} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right) \tag{4}
$$
summing over even and over odd integers:
$$
   f(\alpha, \beta) = \frac{\pi}{2} \frac{ \cos\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) |  \cos \left( \frac{\pi \alpha}{2} \right)} - \frac{\pi}{2} \frac{ \sin\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) |  \sin \left( \frac{\pi \alpha}{2} \right)} = \pi \frac{\sin \left( \alpha \left( \frac{\pi}{2} - \arcsin \cos(\pi \beta) \right) \right)}{ | \sin \pi \beta | \sin(\pi \alpha)}
$$
Now $\frac{\pi}{2} - \arcsin \cos(\pi \beta) = \arccos \cos(\pi \beta) = \pi | \beta |$ for $-1<\beta<1$. Thus, restoring parity, we recover the OP's expression:
$$
    f(\alpha, \beta) = \pi \frac{ \sin(\pi \alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} = \frac{\operatorname{sinc}(\pi \alpha \beta)}{\operatorname{sinc}(\pi \alpha) \operatorname{sinc}(\pi \beta)} \tag{5}
$$
