Fourier-Transformation/Normalverteilung/Dichte/Fakt/Beweis

Beweis

Mit quadratischer Ergänzung ist

{}{\frac {{\mathfrak {t}}^{2}}{2}}-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}={\frac {{\mathfrak {t}}^{2}-2{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}{2}}={\frac {{\left({\mathfrak {t}}-{\mathrm {i} }{\mathfrak {u}}\right)}^{2}+{\mathfrak {u}}^{2}}{2}}\,.

Daher ist mit Fakt

{}{\begin{aligned}{\hat {f}}({\mathfrak {u}})&={\frac {1}{\sqrt {2\pi }}}\int _{\mathbb {R} }e^{-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}e^{\frac {{\mathfrak {t}}^{2}}{2}}d{\mathfrak {t}}\\&={\frac {1}{\sqrt {2\pi }}}\int _{\mathbb {R} }e^{{\frac {{\mathfrak {t}}^{2}}{2}}-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}d{\mathfrak {t}}\\&={\frac {1}{\sqrt {2\pi }}}\int _{\mathbb {R} }e^{\frac {{\left({\mathfrak {t}}-{\mathrm {i} }{\mathfrak {u}}\right)}^{2}+{\mathfrak {u}}^{2}}{2}}d{\mathfrak {t}}\\&={\frac {1}{\sqrt {2\pi }}}e^{\frac {{\mathfrak {u}}^{2}}{2}}\int _{\mathbb {R} }e^{\frac {{\left({\mathfrak {t}}-{\mathrm {i} }{\mathfrak {u}}\right)}^{2}}{2}}d{\mathfrak {t}}\\&=e^{\frac {{\mathfrak {u}}^{2}}{2}},\end{aligned}}

wobei im Integral wegen der Symmetrie der Imaginärteil gleich ${}0$ ist.