Fourier Transform of Gaussian Function

From ProofWiki
Jump to navigation Jump to search

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