Trasformazione di Fourier

$f(x)=\int^{\infty}_{-\infty}{F(k)e^{2\pi ikx}dk}\quad;\quad F(k)=\int^{\infty}_{-\infty}{f(x)e^{-2\pi ikx}dx}$

$F(k)=\mathcal{F}_x[f(x)](k)=\int^{\infty}_{-\infty}{f(x)e^{-2\pi ikx}dx}$

$f(x)=\mathcal{F}^{-1}_k[F(k)](x)=\int^{\infty}_{-\infty}{F(k)e^{2\pi ikx}dk}$

$\mathcal{F}[f(x)]=F(k)\quad;\quad \mathcal{F}[g(x)]=G(k)\;\;\text{ise}$

1.

$\mathcal{F}[af(x)+bg(x)]=a\mathcal{F}[f(x)]+b\mathcal{F}[g(x)]=aF(k)+bF(k)$

2.

$\mathcal{F}_x[1](k)=\int^{\infty}_{-\infty}{f(x)e^{-2\pi ikx}dx}=\delta(k)$

3.

$\begin{align*} \mathcal{F}[\sin(2\pi k_0 x)](k) &= \frac{1}{2}i[\delta(k+k_0)-\delta(k-k_0)] \end{align*}$

4.

$\begin{align*} \mathcal{F}[\delta(x-x_0)](k) &= \int^{\infty}_{-\infty}{\delta(x-x_0) e^{-2\pi ikx}dx}\\ &= e^{-2\pi ikx_0} \end{align*}$

5.

$\begin{align*} \mathcal{F}[e^{-k_0|x|}](k) &= \int^{\infty}_{-\infty}{e^{-k_0|x|} e^{-2\pi ikx}dx}\\ &= \frac{1}{\pi}\frac{k_0}{k^2+k^2_0} \end{align*}$

6.

$\begin{align*} \mathcal{F}[e^{-ax^2}](k) &= \int^{\infty}_{-\infty}{e^{-ax^2} e^{-2\pi ikx}dx}\\ &= \sqrt{\frac{\pi}{a}}e^{-\pi^2k^2/a} \end{align*}$

7.

$\begin{align*} \mathcal{F}[H(x)](k) &= \int^{\infty}_{-\infty}{e^{-2\pi ikx}H(x)dx}\\ &= \frac{1}{2}\left[\delta(k)-\frac{i}{\pi k} \right ] \end{align*}$

8.

$\begin{align*} \mathcal{F}[f'(x)](k) &= 2\pi ik\mathcal{F}[f(x)](k) \end{align*}$

9.

$\begin{align*} \mathcal{F}[f^{(n)}(x)](k) &= (2\pi ik)^n\mathcal{F}[f(x)](k) \end{align*}$