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I am trying to calculate the Fourier transform of $f(t)=Ae^{-i\omega_0 t}$

I'm getting an infinity which is giving me problems. Here are my steps:

$$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{i\omega t}dt$$ $$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty Ae^{-i\omega_0 t}e^{i\omega t}dt$$ $$=\frac{A}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{t(i\omega - i\omega_0)}dt$$ $$=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)} |e^{t(i\omega - i\omega_0)}|_{-\infty}^{\infty}$$ $$=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)}(\infty - 0)$$

Where am I going wrong? Thanks.

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    $\begingroup$ You are going wrong nowhere. The Fourier transform (in the simple-minded setting) is undefined. $\endgroup$ – Fabian Oct 16 '12 at 9:40
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Strictly speakin, $e^{i\omega_0 t}$ does not have a fourier transform, because $e^{i\omega_0 t} \notin L_1$. Observe that $e^{i\omega_0 t}$ contains just a single frequence, $\omega_0$. Thus, to have $$ f(t) =e^{i\omega_0 t} = \int_{-\infty}^\infty F(\omega)e^{i\omega t} d\omega $$ you'd need to have $F(\omega_0) = 1$ and $F(\omega) = 0$ for $\omega \neq \omega_0$. That doesn't work out mathematically, however, because the integral is then just $0$ everywhere.

Actually assigning a fourier transform to $e^{i\omega_0 t}$ works only if you extend the concept of functions to include objects (called distributions) for which the following holds $$ \begin{eqnarray} \delta_{x_0}(x) &=& 0 &\text{ if $x \neq x_0$} \\ \int_{-\infty}^\infty \delta_{x_0}(x) g(x) dx &=& g(x_0) &\text{ for suitable functions $g$ } \end{eqnarray} $$ Note that you should read these more as definitions than as equations. We did, after all, invent new mathematical objects $\delta_{x_0}$ here, so the usual definitions of what it means to evaluate a function (which $\delta_{x_0}$ isn't!), and what it means to integrate don't really apply.

The fourier transform of $f$ is then $F(\omega) = A\delta_{\omega_0}(\omega)$, in a way. The precise definition of what $\delta$ is isn't easy, though - note that you'll have to specify which functions $g$ are suitable. Basically, what you do is to pick a set $X$ of functions as candidates for $g$, called the space of test functions, and then define distributions as linear continuous mappings from $X$ to $\mathbb{R}$. Getting that right requires quite a bit of functional analysis - mostly because it turns out that picking a suitable set $X$ and a topology on $X$ isn't at all trivial. See also http://en.wikipedia.org/wiki/Distribution_%28mathematics%29

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