complex analysis - Contour integration for the improper integral

How to integrate
$\Large \int_{-\infty}^{\infty}\frac{x \sin (7x)}{(x^2-6x+18)} dx$ with contours?

So far I have
$\Large \int_{C}^{}\frac{ze^{7iz}}{(z-(3+3i)) (z-(3-3i))} dz$

With the $$ \large {Res}[f(z),3+3i] = \lim_{z\to3} \ (z-(3+3i))\frac{ze^{7iz}}{(z-(3+3i)) (z-(3-3i))} = \frac{(3+3i)e^{7i(3+3i)}}{(3+3i)-(3-3i))} = \frac{(3+3i)e^{(21i-21)}}{6i}$$

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

$$\sqrt{2}\pi e^{-21}\sin(21+\pi/4)$$