Define a function f(α,β), α∈(−1,1), β∈(−1,1) as

f(α,β)=∫∞0dxxα1+2xcos(πβ)+x2

One can use, for example, the Residue Theorem to show that

f(α,β)=πsin(παβ)sin(πα)sin(πβ)

Clearly, from this latter expression, f(α,β)=f(β,α). My question is, can one see this symmetry directly from the integral expression?

**Answer**

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 α↦−α:

∫∞0xα1+2xcos(πβ)+x2dx=∫10xα1+2xcos(πβ)+x2dx+∫∞1xα1+2xcos(πβ)+x2dx

Make a change of variables x→x−1 in the last integral to obtain:

f(α,β)=∫10xα+x−α1+2xcos(πβ)+x2dx

Now, making a change of variables x=exp(−t) we have:

f(α,β)=∫∞0cosh(αt)cosh(t)+cos(βπ)dt

Using

∫∞0exp(−u(cosht+cosπβ))du=1cosh(t)+cos(βπ)

and the integral representation of the modified Bessel function of the second kind:

∫∞0cosh(αt)exp(−ucosht)dt=Kα(u)

we arrive at a compact representation:

f(α,β)=∫∞0Kα(u)e−ucos(πβ)du

expanding the exponential into series and using ∫∞0unKα(u)du=2n−1Γ(n2+1+α2)Γ(n2+1−α2) we get:

f(α,β)=∞∑n=02n−1n!(−cosπβ)nΓ(n2+1+α2)Γ(n2+1−α2)

summing over even and over odd integers:

f(α,β)=π2cos(αarcsincos(πβ))|sin(πβ)|cos(πα2)−π2sin(αarcsincos(πβ))|sin(πβ)|sin(πα2)=πsin(α(π2−arcsincos(πβ)))|sinπβ|sin(πα)

Now π2−arcsincos(πβ)=arccoscos(πβ)=π|β| for −1<β<1. Thus, restoring parity, we recover the OP's expression:

f(α,β)=πsin(παβ)sin(πα)sin(πβ)=sinc(παβ)sinc(πα)sinc(πβ)

**Attribution***Source : Link , Question Author : Ron Gordon , Answer Author : Sasha*