# Find the fourier transform of $f(t)=e^{-at}$

Firstly, because I've seen many different definitions of the fourier transform with many different variations of the coefficients, my definition is $F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$

I want to find the fourier transform of $f(t)=\begin{bmatrix} 0, t\leq 1 \\e^{-at}, t>1\end{bmatrix}$

So using the definition, it's $$F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt=\frac{1}{2\pi}(\int_{-\infty}^{1}0dt+\int_{1}^{\infty}e^{-at}e^{-i\omega t}dt)=\frac{1}{2\pi}\int_{1}^{\infty}e^{-(a+i\omega)t}dt=\frac{1}{2\pi}[\frac{e^{-(a+i\omega)t}}{-(a+i\omega)}]_{t=1}^{\infty}$$

and here is exactly my problem.

$e^{-\infty}$ approaches zero, that is obvious. But what is $\lim_{t \to \infty}e^{it}$? doesnt the imaginary unit ruin everything? that's the same as $\lim_{t \to \infty} \cos(t)+i\sin(t)$ and the trigonometric functions dont have a defined limit at infinity

• $|e^z| = e^{\operatorname{re}z}$, so in the above, if $a>0$ (which you need for $f$ to be integrable), the quantity goes to zero as $t \to \infty$. – copper.hat Feb 14 '15 at 16:03
• actually i neglected to mention that, but we are given that $a>0$ – Oria Gruber Feb 14 '15 at 16:07
• Then since $e^{-(a+iw)t} = e^{-at} e^{-i wt}$ you see that this $\to 0$ as $t \to \infty$. – copper.hat Feb 14 '15 at 16:18

You're correct that $\lim_{t\to\infty} e^{it}$ doesn't exist. But $\|e^{it}\| \leq 1$ for all values of $t$, and since $\lim_{t\to\infty} e^{-t} = 0$, it follows that $$\lim_{t\to\infty} e^{-(1-i)t} = \lim_{t\to\infty} \underbrace{e^{-t}}_{\to 0} \cdot \underbrace{e^{it}}_{\text{bounded}} = 0.$$
Similarly, $$\lim_{t\to\infty} e^{-(a+i\omega)t} = \lim_{t\to\infty} e^{-at}\cdot e^{-i\omega t} = 0,$$ assuming that $a > 0$.
$$\int_{1}^{M}e^{-(a+i\omega)t}\,dt = \frac{1}{a+i\omega}\left(e^{-(a+i\omega)}-e^{-M(a+i\omega)}\right)\tag{1}$$ Now consider that: $$\left\|e^{-M(a+i\omega)}\right\|=e^{-Ma}$$ so, assuming $a>0$, the last term in the RHS of $(1)$ vanishes when $M\to +\infty$, not because $e^{Mi\omega}$ has a limit, but just because it is bounded regardless of $M$, while $e^{-Ma}$ goes to zero.