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I am trying to obtain the Fourier Transform of:

$$f(t) = \begin{cases} A_0 e^{\frac{-\omega_0t}{2Q}}e^{-i\omega_0t}, &t\ge 0\\ 0, &t < 0 \end{cases}$$

Using the convention:

$$\mathfrak{F}\{f(t)\} = g(\omega)= \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(t)e^{i\omega t}\,dt,$$

$$\mathfrak{F}\{f(t)\} = g(\omega)=\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}e^{\frac{-\omega_0t}{2Q}}e^{-i\omega_0t} e^{i\omega t}\,dt= \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{0} e^{t \left[ {\frac{-\omega_0}{2Q}}+i(\omega-\omega_0)\right]}\,dt.$$

I want to propose the following change of variable:

$$u = t \left[ {\frac{-\omega_0}{2Q}}+i(\omega-\omega_0)\right] ; \, du = \left[ \frac{-\omega_0}{2Q}+i(\omega-\omega_0)\right] dt.$$

So the next step would be:

$$g(\omega)=\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{0} e^{-u}\frac{du}{\frac{-\omega_0}{2Q}+i(\omega-\omega_0)} =\frac{1}{\sqrt{2\pi}} \cdot \frac{1}{\frac{-\omega_0}{2Q}+i(\omega-\omega_0)}\int^{+\infty}_{0} e^{-u}\,du,$$

$$\int^{+\infty}_{-\infty} e^{-u}\,du = -e^{-u}\big|^{\infty}_{0}=-e^{\infty}+e^{0}=0+1=1$$

So $$g(\omega)=\frac{1}{\sqrt{2\pi}} \cdot \frac{1}{\frac{-\omega_0}{2Q}+i(\omega-\omega_0)}.$$

Can anybody help me to confirm this result (and procedure), my doubt is if I can make the change of variables proposed.

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  • $\begingroup$ The function $f(t)$ has exponential growth for $t<0$. Are you certain that $f$ is correctly given? $\endgroup$ – Mark Viola Apr 20 '17 at 4:26
  • $\begingroup$ I forgot to mention that $f(t) = 0 $for $t < 0$, this will change the limits of the integral, I will edit this post $ $\endgroup$ – IvanMartinez Apr 20 '17 at 4:29
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There is a sign error in the result given in the OP.

Let $f(t)=e^{-\omega_0 t/2Q}e^{-i\omega_0 t}u(t)$. Then, we have

$$\begin{align} F(\omega)&=\frac{1}{\sqrt{2\pi }}\int_0^\infty e^{-\omega_0 t/2Q}e^{-i\omega_0 t}e^{i\omega t}\,dt\\\\ &=\frac{1}{\sqrt{2\pi }}\int_0^\infty e^{-\left(\omega_0 /2Q-i(\omega-\omega_0)\right)t}\,dt\\\\ &=\frac{1}{\sqrt{2\pi }}\,\,\frac{1}{\omega_0 /2Q-i(\omega-\omega_0)} \end{align}$$

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