show that ∫∞−∞(sinx)(x2+a2)x(x2+b2)dx=π(a2+e−b(b2−a2))b2
for every a,b>0
thanks for all
Answer
Consider the contour integral
∮Cdz(z2+a2)eizz(z2+b2)
where C is the semicircular contour of radius R in the upper half-plane, with an additional semicircular contour of radius ϵ centered at the origin, jutting into the upper half-plane.
The contour integral is equal to
∫−ϵ−Rdx(x2+a2)eixx(x2+b2)+iϵ∫0−πdϕeiϕ(a2+ϵ2ei2ϕ)eiϵeiϕϵeiϕ(b2+ϵ2ei2ϕ)+∫Rϵdx(x2+a2)eixx(x2+b2)+iR∫π0dθeiθ(a2+R2ei2θ)eiReiθReiθ(b2+R2ei2θ)
We take the limit as R→∞ and ϵ→0 and get
PV∫∞−∞dx(x2+a2)eixx(x2+b2)−iπa2b2
Note that the magnitude of fourth integral vanishes as
2∫π/20e−Rsinθ≤2∫π/20e−2Rθ/π≤πR
as R→∞. By the residue theorem, the contour integral is equal to i2π times the residue of the pole at z=ib. Therefore
PV∫∞−∞dx(x2+a2)eixx(x2+b2)=iπa2b2−i2π(a2−b2)e−b2b2
Similarly,
PV∫∞−∞dx(x2+a2)e−ixx(x2+b2)=−iπa2b2+i2π(a2−b2)e−b2b2
Therefore,
∫∞−∞dx(x2+a2)sinxx(x2+b2)=πa2b2−π(a2−b2)e−bb2
which is equivalent to the stated result.
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