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Fourier transform
The Fourier transform $F(s)$ of a function $f(t)$ is defined as follows:
$F(s)=\frac{1}{\sqrt{2\pi}}\int_{{\infty}}^{{\infty}}e^{{ist}}f(t)dt.$ 
The Fourier transform exists if $f$ is Lebesgue integrable on the whole real axis.
If $f$ is Lebesgue integrable and can be divided into a finite number of continuous, monotone functions and at every point both onesided limits exist, the Fourier transform can be inverted:
$f(t)=\frac{1}{\sqrt{2\pi}}\int_{{\infty}}^{{\infty}}e^{{ist}}F(s)ds.$ 
Sometimes the Fourier transform is also defined without the factor $\frac{1}{\sqrt{2\pi}}$ in one direction, but therefore giving the transform into the other direction a factor $\frac{1}{2\pi}$. So when looking a transform up in a table you should find out how it is defined in that table.
The Fourier transform has some important properties, that can be used when solving differential equations. We denote the Fourier transform of $f$ with respect to $t$ in terms of $s$ by $\mathcal{F}_{t}(f)$.

$\mathcal{F}_{t}(af+bg)=a\mathcal{F}_{t}(f)+b\mathcal{F}_{t}(g),$
where $a$ and $b$ are constants. 
$\mathcal{F}_{t}\left(\frac{\partial}{\partial t}f\right)=is\mathcal{F}_{t}(f).$

$\mathcal{F}_{t}\left(\frac{\partial}{\partial x}f\right)=\frac{\partial}{% \partial x}\mathcal{F}_{t}(f).$

We define the bilateral convolution of two functions $f_{1}$ and $f_{2}$ as:
$(f_{1}\ast f_{2})(t):=\frac{1}{\sqrt{2\pi}}\int_{{\infty}}^{{\infty}}f_{1}(% \tau)f_{2}(t\tau)d\tau.$ Then the following equation holds:
$\mathcal{F}_{t}((f_{1}\ast f_{2})(t))=\mathcal{F}_{t}(f_{1})\cdot\mathcal{F}_{% t}(f_{2}).$
If $f(t)$ is some signal (maybe a wave) then the frequency domain of $f$ is given as $\mathcal{F}_{t}(f)$. Rayleigh’s theorem states that then the energy $E$ carried by the signal $f$ given by:
$E=\int_{{\infty}}^{{\infty}}f(t)^{2}dt$ 
can also be expressed as:
$E=\int_{{\infty}}^{{\infty}}\mathcal{F}_{t}(f)(s)^{2}ds.$ 
In general we have:
$\int_{{\infty}}^{{\infty}}f(t)^{2}dt=\int_{{\infty}}^{{\infty}}\mathcal{F% }_{t}(f)(s)^{2}ds,$ 
also known as the first Parseval theorem.
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Comments
Better form
I think it might be better to define the transform as for f, let
F(s) = int( f(x)*e^(2*pi*i*s*x) dx, x=oo,oo).
and then correspondingly,
f(x) = int( F(s)*e^(2*pi*s*x) ds, s=oo,oo).
Re: Better form
I have never seen it in that form, does it work?
"Do not meddle in the affairs of wizards for they are subtle and quick to anger."
Re: Better form
I hope so, since that's the form we used my entire class last term. It seems a bit more symmetrical to me.
Re: Better form
The form without the oneonroot(2*pi) is usually used for beginners (as it is easier) or statisticians as their characteristic functions of distributions come out nicely.
The form with the oneonroot(2*pi) is a bit nicer mathematically as it makes the fourier transform an isometry.
Putting the 2*pi in the exponent also makes it an isometry, though it does make some of the properties (eg, translation) seem a little strange. However, the payoff comes with any sort of representation thoery. Also see Poisson's summation formula.