Fourier-Transformation/Rechtsseitig abfallende Exponentialfunktion/Beispiel

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

${}f({\mathfrak {t}})={\begin{cases}e^{-a{\mathfrak {t}}}{\text{ für }}{\mathfrak {t}}\geq 0\\0{\text{ sonst}}\,.\ \end{cases}}\,$

Es ist

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