(Originally posed in Math.SE in Jan 2013. Received no complete answers as of yet.)

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?

  • 2
    $\begingroup$ the final answer on MSE, which rewrites the integral identically as $f(\alpha,\beta)=-(\sin\pi\alpha\beta)(\sin\pi\beta)^{-1}\int_0^\infty dx\,x^\alpha/(1+x)$ goes a long way towards demonstrating the symmetry (because the remaining integral evaluates to $-\pi/\sin\pi\alpha$); but you probably want to avoid any evaluation at all, right? $\endgroup$ – Carlo Beenakker Jun 21 '16 at 14:44
  • $\begingroup$ @CarloBeenakker: that is correct - we should be able to see why the evaluation leads to a symmetric function without any evaluation of the integral. Or so I hope. $\endgroup$ – Ron Gordon Jun 21 '16 at 14:45
  • 1
    $\begingroup$ It seems tempting to write the integrand in terms of a Poisson kernel: en.wikipedia.org/wiki/Poisson_kernel#On_the_unit_disc $\endgroup$ – Steve Huntsman Jun 22 '16 at 12:30
  • $\begingroup$ have a look at this posting --- there is some integration involved, but only elementary integrals, no contour integration --- does it qualify as an answer? $\endgroup$ – Carlo Beenakker Jun 22 '16 at 18:12

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