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How do you derive the Fourier transform of $H(t)e^{-at}$?

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} H(\omega) \, e^{-a\omega+i\omega t}\, \mathrm{d}\omega$$

I tried $\frac{1}{2\pi} \int_{-\infty}^{\infty} H(\omega) \, e^{-a\omega+i\omega t}\, \mathrm{d}\omega$. I split it into three integrals ($-\infty \lt 0$, $0 \,\text{to} \, 0$, $0 \ge \infty$), with the first two becoming 0 and the last remaining.

Defining $H(t) = $0 when $t \le 0$ and $1$ when greater or equal to $0$, I got = $$\left(\frac{1}{2\pi}\right)\left(\frac{1}{a+i\omega}\right) + \left(\frac{1}{2\pi}\right)\left(\frac{\infty - 1}{a+i\omega}\right)$$

BLAZE
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Assuming $a > 0$, and $\omega$ is real, then $|e^{(-a-i\omega)t}|=e^{-at}\rightarrow 0$ as $t\rightarrow\infty$. Therefore, \begin{align} \int_{-\infty}^{\infty}(H(t)e^{-at})e^{-i\omega t}dt &=\int_{0}^{\infty}e^{-at}e^{-i\omega t}dt\\ &=\int_{0}^{\infty}e^{(-a-i\omega)t}dt\\ &=\frac{1}{-a-i\omega}e^{(-a-i\omega)t}|_{t=0}^{\infty}\\ &= \frac{1}{a+i\omega}. \end{align} The way people deal with the factors of $2\pi$ varies, and I'll let you sort that out.

Disintegrating By Parts
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  • Indeed... and/but there is still the issue of justifying that change of variables, since $a-i\omega$ is not real. E.g., by the identity principle from complex analysis, first treating the case that $s=a-i\omega$ is real... But, yes, this is secondary to the main idea as illustrated. – paul garrett Oct 21 '15 at 21:28
  • @paulgarrett : Which is essentially accomplished by proving that $\frac{d}{dt}\frac{1}{-a-i\omega}e^{(-a-i\omega)t}=e^{(-a-i\omega)t}$. – Disintegrating By Parts Oct 21 '15 at 21:33
  • Yes, if one doesn't worry about how the limits change in the definite integral. Again, yes, certainly secondary to seeing what should happen. – paul garrett Oct 22 '15 at 14:39
  • @paulgarrett : I think I have misunderstood your question. Are you referring to the truncation of the integral due to the presence of the Heaviside step function $H$? – Disintegrating By Parts Oct 22 '15 at 15:12
  • No, only in effect referring to the variant in which one says "change variables in the integral", but replacing a real variable by a complex multiple of it changes the contour over which one is integrating, etc. I imagined that the student might care about that option, given the context in which the question makes sense.... or maybe I'm just clouding the issue... – paul garrett Oct 22 '15 at 15:25
  • @paulgarrett : That didn't happen here, did it? – Disintegrating By Parts Oct 22 '15 at 15:53
  • That variant is not here now, but I think the variant (=change-of-variables) either appeared in a now-vanished comment here, or on a very similar question somewhere around here. – paul garrett Oct 22 '15 at 16:04