Fourier-Transformation/Beidseitig abfallende Exponentialfunktion/Beispiel

Es sei ${}a>0$ fixiert. Wir betrachten die Funktion

${}f({\mathfrak {t}})=e^{-a\vert {\mathfrak {t}}\vert }\,.$

Es ist unter Verwendung von Beispiel

${}{\begin{aligned}{\hat {f}}({\mathfrak {u}})&={\frac {1}{\sqrt {2\pi }}}\int _{\mathbb {R} }e^{-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}e^{-a\vert {\mathfrak {t}}\vert }d{\mathfrak {t}}\\&={\frac {1}{\sqrt {2\pi }}}\int _{0}^{\infty }e^{-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}e^{-a{\mathfrak {t}}}d{\mathfrak {t}}+{\frac {1}{\sqrt {2\pi }}}\int _{0}^{-\infty }e^{-{\mathrm {i} }{\mathfrak {u}}{\mathfrak {t}}}e^{a{\mathfrak {t}}}d{\mathfrak {t}}\\&={\frac {1}{\sqrt {2\pi }}}\cdot {\frac {1}{a+{\mathrm {i} }{\mathfrak {u}}}}-{\frac {1}{\sqrt {2\pi }}}\int _{0}^{\infty }e^{{\mathrm {i} }{\mathfrak {u}}s}e^{-as}ds\\&={\frac {1}{\sqrt {2\pi }}}\cdot {\frac {1}{a+{\mathrm {i} }{\mathfrak {u}}}}+{\frac {1}{\sqrt {2\pi }}}\cdot {\frac {1}{a-{\mathrm {i} }{\mathfrak {u}}}}\\&={\frac {1}{\sqrt {2\pi }}}\cdot {\frac {2a}{a^{2}+{\mathfrak {u}}^{2}}}\\&={\frac {\sqrt {2}}{\sqrt {\pi }}}\cdot {\frac {a}{a^{2}+{\mathfrak {u}}^{2}}}.\end{aligned}}$