Fourier Transform of Gaussian Function
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Theorem
Let $\map f x$ be defined as $\sqrt \pi$ times the Gaussian probability density function where $\mu = 0$ and $\sigma = \dfrac {\sqrt 2} 2$:
\(\ds \map f x\) | \(=\) | \(\ds \dfrac {\sqrt {\pi} } {\dfrac {\sqrt 2} 2 \sqrt {2 \pi} } \map \exp {-\dfrac {\paren {x - 0}^2} {2 \paren {\dfrac {\sqrt 2} 2 }^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{-x^2}\) |
Then:
- $\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s }^2}$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.
Proof
By the definition of a Fourier transform:
\(\ds \map {\hat f} s\) | \(=\) | \(\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-x^2} \rd x\) |
Taking the derivative with respect to $s$, we have:
\(\ds \dfrac \d {\d s} \map {\hat f} s\) | \(=\) | \(\ds \dfrac \d {\d s} \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \pi i \int_{-\infty}^\infty e^{-2 \pi i x s} x e^{-x^2} \rd x\) |
Integrating by parts, we have:
Let $u = e^{-2 \pi i x s}$ and $\rd v = x e^{-x^2} \rd x$
Then:
- $\rd u = -2 \pi i s e^{-2 \pi i x s} \rd x$ and $v = -\dfrac 1 2 e^{-x^2}$
Hence:
\(\ds \dfrac \d {\d s} \map {\hat f} s\) | \(=\) | \(\ds -2 \pi i \paren {\paren {\lim_{\gamma \mathop \to \infty} \intlimits {-\dfrac 1 2 e^{-2 \pi i x s} e^{-x^2} } {-\gamma} {+\gamma} } - \int_{-\infty}^\infty \paren {-\dfrac 1 2 e^{-x^2} } \paren {-2 \pi i s e^{-2 \pi i x s} } \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \pi i \paren {0 - \int_{-\infty}^\infty \paren {-\dfrac 1 2 e^{-x^2} } \paren {-2 \pi i s e^{-2 \pi i x s} } \rd x}\) | first part $\paren {u v}$ sums to zero | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \pi i \paren {-\pi i \int_{-\infty}^\infty \paren {e^{-x^2} } \paren { s e^{-2 \pi i x s} } \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi^2 i^2 \int_{-\infty}^\infty \paren {e^{-x^2} } \paren {s e^{-2 \pi i x s} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \pi^2 \int_{-\infty}^\infty \paren {e^{-x^2} } \paren {s e^{-2 \pi i x s} } \rd x\) | $i^2 = -1$ |
We now have the following:
\(\ds \frac \d {\d s} \map {\hat f} s\) | \(=\) | \(\ds \int_{-\infty}^\infty \paren {-2 \pi^2 s e^{-x^2} } e^{-2 \pi i x s} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \pi^2 s \map {\hat f} s\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \d {\map {\hat f} s}\) | \(=\) | \(\ds -2 \pi^2 s \map {\hat f} s \rd s\) | |||||||||||
\(\ds \dfrac {\map \d {\map {\hat f} s } } {\map {\hat f} s}\) | \(=\) | \(\ds -2 \pi^2 s \rd s\) | Solution to Separable Differential Equation | |||||||||||
\(\ds \int \dfrac {\map \d {\map {\hat f} s } } {\map {\hat f} s}\) | \(=\) | \(\ds \int -2 \pi^2 s \rd s\) | integrating | |||||||||||
\(\ds \map \ln {\map {\hat f} s}\) | \(=\) | \(\ds -\pi^2 s^2 + C\) | Primitive of Function under its Derivative, Primitive of Power | |||||||||||
\(\ds \map {\hat f} s\) | \(=\) | \(\ds A e^{-\pi^2 s^2}\) | Exponential of Natural Logarithm: $A = e^C$ |
We solve for $A$ by setting $s = 0$:
\(\ds \map {\hat f} 0\) | \(=\) | \(\ds \int_{-\infty}^\infty e^{-2 \pi i x 0} e^{-x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{-\infty}^\infty e^{-x^2} \rd x\) | Exponential of Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt \pi\) | Gaussian Integral |
Therefore:
- $\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s}^2}$
$\blacksquare$
Sources
- Weisstein, Eric W. "Fourier Transform." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/FourierTransform.html