Sunday, 28 April 2019

complex analysis - Contour integration for the improper integral



How to integrate
xsin(7x)(x26x+18)dx with contours?




So far I have
Cze7iz(z(3+3i))(z(33i))dz



With the Res[f(z),3+3i]=limz3 (z(3+3i))ze7iz(z(3+3i))(z(33i))=(3+3i)e7i(3+3i)(3+3i)(33i))=(3+3i)e(21i21)6i



and Res[f(z),33i]=limz3 (z(33i))ze7iz(z(3+3i))(z(33i))=(33i)e7i(33i)((33i)(3+3i))=(33i)e(21i+21)6i





For the upper half plane 2πiRes[f(z),3+3i]=π(3+3i)e(21i21)3=π(1+i)e(21i21)




and for the upper lower half plane and 2πiRes[f(z),33i]=π(33i)e(21i+21)3=π(1i)e(21i+21)



so
xcos(7x)(x26x+18)dx+ixsin(7x)(x26x+18)dx+Cze7iz(z(3+3i))(z(33i))dz=π(1+i)e(21i21)+π(1i)e(21i+21)



and equating the imaginary parts



xsin(7x)(x26x+18)dx=iπ(e21i21e21i+21)


Answer




When transforming to the z plane we have



sin7xsin7zei7z



However,



sinz=Im{ei7z}



so that xsin(7x)(x26x+18)dx becomes




Im(Czei7zz26z+18dz)



where the contour C is composed of the real axis plus the "infinite semi-circle" that encloses the upper half plane. You use Jordan's lemma to show that the integration over the semi-circle does not contribute to the closed-contour integration. Then, the integral of interest is equal to 2πi times the imaginary part of the residue of



zei7zz26z+18



in the upper-half plane. Thus,



xsin(7x)(x26x+18)dx=Im(2πiRes(zei7zz26z+18))




Working out the details, the integral is



2πe21sin(21+π/4)


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