How to integrate
∫∞−∞xsin(7x)(x2−6x+18)dx with contours?
So far I have
∫Cze7iz(z−(3+3i))(z−(3−3i))dz
With the Res[f(z),3+3i]=limz→3 (z−(3+3i))ze7iz(z−(3+3i))(z−(3−3i))=(3+3i)e7i(3+3i)(3+3i)−(3−3i))=(3+3i)e(21i−21)6i
and Res[f(z),3−3i]=limz→3 (z−(3−3i))ze7iz(z−(3+3i))(z−(3−3i))=(3−3i)e7i(3−3i)((3−3i)−(3+3i))=(3−3i)e(21i+21)−6i
For the upper half plane 2πiRes[f(z),3+3i]=π(3+3i)e(21i−21)3=π(1+i)e(21i−21)
and for the upper lower half plane and −2πiRes[f(z),3−3i]=π(3−3i)e(21i+21)3=π(1−i)e(21i+21)
so
∫∞−∞xcos(7x)(x2−6x+18)dx+i∫∞−∞xsin(7x)(x2−6x+18)dx+∫Cze7iz(z−(3+3i))(z−(3−3i))dz=π(1+i)e(21i−21)+π(1−i)e(21i+21)
and equating the imaginary parts
∫∞−∞xsin(7x)(x2−6x+18)dx=iπ(e21i−21−e21i+21)
Answer
When transforming to the z plane we have
sin7x→sin7z≠ei7z
However,
sinz=Im{ei7z}
so that ∫∞−∞xsin(7x)(x2−6x+18)dx becomes
Im(∮Czei7zz2−6z+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
zei7zz2−6z+18
in the upper-half plane. Thus,
∫∞−∞xsin(7x)(x2−6x+18)dx=Im(2πiRes(zei7zz2−6z+18))
Working out the details, the integral is
√2πe−21sin(21+π/4)
No comments:
Post a Comment