We have
$$f(t)=\frac{1}{t^2+6t+13}\tag{1} \qquad \& \qquad \hat{f}(\omega)=\frac{\pi}{2}\cdot e^{3\omega i}e^{2|\omega|}$$
whereas $\hat{f}$ denotes the Fourier transform.
We want to calculate the following integral using this result:
$$K=\int_{-\infty}^\infty \frac{\sin(3t)}{t^2+6t+13}dt \tag{2}$$
Solution:
We know that
$$\mathscr{F}[f] (t)=\hat{f}(\omega)=\int_{-\infty}^\infty \frac{e^{-i\omega t}}{t^2+6t+13}dt=\int_{-\infty}^\infty \frac{\cos(-\omega t)+i\sin(-\omega t)}{t^2+6t+13}dt \tag{3}$$
so we see that
$$K\overset{!}{=} \operatorname{Im}\hat{f}(\omega)\overset{!}{=}\int_{-\infty}^\infty \frac{\sin(-\omega t)}{t^2+6t+13}dt \quad \Rightarrow \quad \omega=-3 \tag{4}$$
so we get
$$K= \operatorname{Im}\hat{f}(-3)=\frac{\pi}{2}\cdot e^{2|\omega|}\cdot \sin(3\omega t)|_{\omega=-3}=\frac{\pi}{2}e^6\sin(-9t) \tag{5}$$
Question: Sadly I don't have any solution for this and I'm not sure if my argumentation holds.
Best Answer
You are on the right track, but I'll point out two things. First, it should be $$ \widehat f(\omega)=\int_{-\infty}^\infty \frac{e^{-i\omega t}}{t^2+6t+13}\mathrm d t=\frac\pi 2 e^{3\omega i}e^{\color{red}{- 2}|\omega|}. $$ And in the last step, $$ K =\Im \widehat f(-3) =\frac\pi 2 \sin(3\omega)e^{-2|\omega|}\Big]_{\omega=-3}=-\frac\pi 2 \sin(9)e^{-6}.$$ (There's no $t$ in the final result.)