Let | $f(x)\stackrel{F.T.}{\longrightarrow}F(q)$, |

and | $g(x)\stackrel{F.T.}{\longrightarrow}G(q).$ |

Then | $f(x)+g(x)\stackrel{F.T.}{\longrightarrow}F(q)+G(q),$ |

*i.e.* *the FT of a sum is the sum the individual FTs*, which is what we might have
expected anyway.

Here's the proof:

$\mathcal{FT}(f(x)+g(x))$ | $=\int(f(x)+g(x)){\rm e}^{-{\rm i}qx}{\rm d}x$ |

$=\frac{1}{2\pi}\int f(x){\rm e}^{-{\rm i}qx}{\rm d}x+\frac{1}{2\pi}\int g(x){\rm e}^{-{\rm i}qx}{\rm d}x$ | |

$=F(q)+G(q).$ |

Unfortunately, multiplication doesn't quite work out so straightforward.

Note that
$f(x)\cdot g(x)\color{red}{\xcancel{\color{black}{\stackrel{F.T.}{\longrightarrow}}}}F(q)\cdot G(q)\qquad$!

Instead, the product of two Fourier transforms is the *convolution* of the two originating
functions:
$$f(x)\ast g(x)\stackrel{F.T.}{\longrightarrow}F(q)\cdot G(q)\qquad.$$

The convolution of $f(x)$ and $g(x)$ is defined as
$$h(x_0):=f(x)\ast g(x)=\int_{-\infty}^{+\infty}f(x)\cdot g(x_0-x){\rm d}x\qquad,$$
*i.e.* the value of the convolution is evaluated separately for each consecutive point
$x_0$ by integration with respect to $x$. The integrand is the product of
one of the functions at $x$ and the other function at *the distance* between $x$
and the fixed value of $x_0$ for which the convolution is to be evaluated.

To visualise this formalism, think of two overlapping spectral lines: If the two lines are far apart, there is little overlap. However, if the two lines come closer together, the overlap increases. The amount of overlap, as seen from line 1, depends on the amplitude of line 1 itself and the distance between lines 1 and 2.

Fourier transforms are based on a superposition of periodic functions. When shifting the range of
a periodic function along the *x* axis, we change the phase of the sine wave. Accordingly,
a phase factor (complex exponential) is included in the Fourier transform:

Again, let | $f(x)\stackrel{F.T.}{\longrightarrow}F(q)\qquad.$ |

To determine the FT of the function shifted by $a$: | $$FT(f(x-a))=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(x-a){\rm e}^{-{\rm i}qx}{\rm d}x\qquad.$$ |

Make $(x-a)$ appear in the exponential: | $$=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(x-a){\rm e}^{-{\rm i}q(x-a)}{\rm e}^{-{\rm i}qa}{\rm d}x\qquad.$$ |

If the differential were ${\rm d}(x-a)$ rather than just ${\rm d}x$, the integral would contain the Fourier transform of $f(x-a)$. To achieve this, we need to shift integration limits.

To figure this out, consider the sum | $$\sum_{n=1}^{5}n=1+2+3+4+5\qquad.$$ |

This is the same as the sum | $$\sum_{n=0}^{4}n+1\qquad,$$ |

or even | $$\sum_{n=1+a}^{5+a}n-a\qquad.$$ |

We can shift the limits by $a$ if we compensate that in the formula of the sum.

The same trick works with integrals because they are just the continuous analogue of sums.

In order to replace ${\rm d}x$ with ${\rm d}(x-a)$, we need to shift the lower and upper integration limits by $+a$.

Luckily, infinity plus $a$ is still infinity, so it doesn't make a difference: | $$\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(x-a){\rm e}^{-{\rm i}q(x-a)}{\rm e}^{-{\rm i}qa}{\rm d}(x-a)\qquad.$$ |

Therefore, | $$f(x-a)\stackrel{F.T.}{\longrightarrow}F(q){\rm e}^{-{\rm i}qa}\qquad.$$ |

To conclude the Fourier section, we'll do some practical transforming.