Exponentialfunktionen/Frequenzen/01/Orthonormalsystem/Fakt/Beweis/Aufgabe/Lösung

Es ist

{}{\begin{aligned}\left\langle f_{m},f_{n}\right\rangle &=\int _{0}^{1}f_{m}{\overline {f_{n}}}dx\\&=\int _{0}^{1}e^{2\pi {\mathrm {i} }mx}{\overline {e^{2\pi {\mathrm {i} }nx}}}dx\\&=\int _{0}^{1}e^{2\pi {\mathrm {i} }mx}e^{-2\pi {\mathrm {i} }nx}dx\\&=\int _{0}^{1}e^{2\pi {\mathrm {i} }(m-n)x}dx.\end{aligned}}

Bei ${}m=n$ ist dies

{}\int _{0}^{1}e^{0}dx=\int _{0}^{1}1dx=1\,.

Bei ${}m\neq n$ ist dies

{}{\begin{aligned}\int _{0}^{1}e^{2\pi {\mathrm {i} }(m-n)x}dx&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(e^{2\pi {\mathrm {i} }(m-n)y}\right)}_{0}^{1}\\&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(e^{2\pi {\mathrm {i} }(m-n)}-e^{0}\right)}\\&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(1-1\right)}\\&=0.\end{aligned}}