Interpretation of Fourier Transform I know that the idea of the Fourier Transform is to break a function into a sum of trigonometric functions. Consider the following function: $$ f_{\alpha}(t) = e^{-\alpha|t|}$$
The Fourier Transform of this is $$\tilde{f}_{\alpha}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i \omega t} e^{-\alpha |t|} \ dt  $$
$$ = \frac{\alpha}{\pi(\alpha^{2}+\omega^{2})}$$
What precisely does this mean? How does did relate to the question of breaking a function into a sum of trigonometric functions?
 A: I had this same question when learning about the Fourier transform on the real line. The Fourier transform on the circle (or some interval) is very clearly motivated. It breaks a function up into periodic pieces so it's easier to handle. But it's hard to see what the Fourier transform on the real line is doing.
The only thing I could find that fully answered my questions is the article "The Fourier Transform" (no. III.27 in the "Princeton Companion to Mathematics") by Terence Tao; a preprint of the article is available on his website. It's excellent.
A: The value of the Fourier transform at a given frequency (i.e., $\omega$) is simply the contribution of that frequency to the signal. If you consider something basic, say a sine wave (which has just a single frequency), then $\tilde{f}(\omega)$ will be zero everywhere except where $\omega$ is equal to the frequency of the sine wave, at which point $\tilde{f}(\omega)$ will have a magnitude equal to the amplitude of the wave and a phase equal to its argument at $t=0$.
The signal in your example, however, is made up of a continuum of frequencies and $\tilde{f}(\omega)$ just tells you the contribution of each.
A: I brought an extensive answer to a very similar question over here and hope that it would halp you. The function there is targeting a discrete inverse Fourier but is general enough to also cover your question that is quite generic.
How to interpret Fourier Transform result?
A: The idea of the Fourier series is that every periodic function can be decomposed into an infinite series of sines and cosines. 
Fourier transform is generalization of this result.Any function $F$ (satisfying some conditions) can be written in the form 
$$F(x)= \int_{- \infty}^{\infty}\bar F(t)e^{2\pi f i  x}df$$
You can interpret this above integral as "decomposing" the function $F$ in terms of $e^{2 \pi f i  x}$. Here $\bar F$ is the fourier transform of $f$.So, the value of  $\bar F$ at $f$ gives the "contribution" of $e^{2 \pi  f i x}$ in this integral. So, Fourier transform is analogous to Fourier coefficients in a Fourier series.
