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:

$$K=\int_{-\infty}^\infty \frac{\sin(3t)}{t^2+6t+13}dt \tag{2}$$


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.

  • $\begingroup$ Das ist doch Aufgabe 13.3d) iv. von Komplexe Analysis Serie 13 ! Freut mich einen Mitstudent auf dieser Website zu finden :D $\endgroup$ – Ryukyu Feb 3 '19 at 22:14

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.)

  • $\begingroup$ Ah yeah, of course there shouldn't be a $t$ in the solution. Thanks! Also didn't had any solution for the Fourier transform, thanks. $\endgroup$ – xotix Feb 3 '19 at 15:58

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