Wednesday, 1 April 2015

calculus - show that intinftyinftyfrac(sinx)(x2+a2)x(x2+b2)dx=fracpi(a2+eb(b2a2))b2



show that (sinx)(x2+a2)x(x2+b2)dx=π(a2+eb(b2a2))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.



enter image description here



The contour integral is equal to




ϵRdx(x2+a2)eixx(x2+b2)+iϵ0πdϕeiϕ(a2+ϵ2ei2ϕ)eiϵeiϕϵeiϕ(b2+ϵ2ei2ϕ)+Rϵdx(x2+a2)eixx(x2+b2)+iRπ0dθeiθ(a2+R2ei2θ)eiReiθReiθ(b2+R2ei2θ)



We take the limit as R and ϵ0 and get



PVdx(x2+a2)eixx(x2+b2)iπa2b2



Note that the magnitude of fourth integral vanishes as



2π/20eRsinθ2π/20e2Rθ/ππR




as R. By the residue theorem, the contour integral is equal to i2π times the residue of the pole at z=ib. Therefore



PVdx(x2+a2)eixx(x2+b2)=iπa2b2i2π(a2b2)eb2b2



Similarly,



PVdx(x2+a2)eixx(x2+b2)=iπa2b2+i2π(a2b2)eb2b2



Therefore,




dx(x2+a2)sinxx(x2+b2)=πa2b2π(a2b2)ebb2



which is equivalent to the stated result.


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