Symmetry of function defined by integral

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 αα:
0xα1+2xcos(πβ)+x2dx=10xα1+2xcos(πβ)+x2dx+1xα1+2xcos(πβ)+x2dx
Make a change of variables xx1 in the last integral to obtain:
f(α,β)=10xα+xα1+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)eucos(πβ)du
expanding the exponential into series and using 0unKα(u)du=2n1Γ(n2+1+α2)Γ(n2+1α2) we get:
f(α,β)=n=02n1n!(cosπβ)nΓ(n2+1+α2)Γ(n2+1α2)
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(π2arcsincos(πβ)))|sinπβ|sin(πα)
Now π2arcsincos(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

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