Pro Fourierovu transformaci Diracovy delta funkce platí
a inverzně
Zavedeme-li sumu delta-funkcí přes všechna celá čísla tak, že
potom
neboť
což je ovšem to samé, jako
![{\displaystyle {\mathcal {F}}[\delta (t)]={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\delta (t-a)e^{-i\omega t}\,\mathrm {d} t={\frac {1}{\sqrt {2\pi }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eed6da5005aa412c628b1980f739ffd477276425)
![{\displaystyle {\mathcal {F}}^{-1}\left[{\frac {1}{\sqrt {2\pi }}}\right]=\delta (t)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{i\omega t}\,\mathrm {d} \omega ={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{i\omega t}\,\mathrm {d} \omega }](https://wikimedia.org/api/rest_v1/media/math/render/svg/136d35673c43665367a513e5d7ca65cd3377b5c9)
![{\displaystyle {\mathcal {F}}[\Sigma (t)]=\Sigma (\omega ),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94fc5b3e168941451a9d13a8ad7a7d8362d54752)
![{\displaystyle {\mathcal {F}}[\Sigma (t)]={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\sum _{k=-\infty }^{\infty }\delta (t-k)e^{-i\omega t}\,\mathrm {d} t={\frac {1}{\sqrt {2\pi }}}\sum _{k=-\infty }^{\infty }e^{-i\omega k},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e24f7d69095790210541dadcbcc9023154a060b)
![{\displaystyle {\mathcal {F}}\left[{\frac {1}{\sqrt {2\pi }}}\sum _{k=-\infty }^{\infty }e^{-itk}\right]={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\sum _{k=-\infty }^{\infty }e^{-it(k+\omega )}\,\mathrm {d} t=\sum _{k=-\infty }^{\infty }e^{-i\omega k},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19a9554eb32402071dab0d8f425c42554f12cfcd)
