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]=lim
and \large {Res}[f(z),3-3i] = \lim_{z\to3} \ (z-(3-3i))\frac{ze^{7iz}}{(z-(3+3i)) (z-(3-3i))} = \frac{(3-3i)e^{7i(3-3i)}}{((3-3i)-(3+3i))} = \frac{(3-3i)e^{(21i+21)}}{-6i}
For the upper half plane \large 2\pi i{Res}[f(z),3+3i] = \pi \frac{(3+3i)e^{(21i-21)}}{3} = \pi(1+i)e^{(21i-21)}
and for the upper lower half plane and \large -2\pi i{Res}[f(z),3-3i] = \pi \frac{(3-3i)e^{(21i+21)}}{3} = \pi(1-i)e^{(21i+21)}
so
\int_{-\infty}^{\infty}\frac{x \cos (7x)}{(x^2-6x+18)} dx + i\int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx + \int_{C}^{}\frac{ze^{7iz}}{(z-(3+3i)) (z-(3-3i))} dz = \pi(1+i)e^{(21i-21)} + \pi(1-i)e^{(21i+21)}
and equating the imaginary parts
\int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx = i\pi(e^{21i-21} - e^{21i+21})
Answer
When transforming to the z plane we have
\sin 7x \to \sin 7z \ne e^{i7z}
However,
\sin z =\text{Im}\{e^{i7z}\}
so that \int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx becomes
\text{Im}\left(\oint_C \frac{z e^{i7z}}{z^2-6z+18}dz\right)
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 2\pi i times the imaginary part of the residue of
\frac{z e^{i7z}}{z^2-6z+18}
in the upper-half plane. Thus,
\int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx=\text{Im}\left(2\pi i \text{Res}\left( \frac{z e^{i7z}}{z^2-6z+18}\right) \right)
Working out the details, the integral is
\sqrt{2}\pi e^{-21}\sin(21+\pi/4)
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