prove-that-0-e-x-2-n-0-e-n-2-

Question Number 30441 by abdo imad last updated on 22/Feb/18

p r o v e t h a t \int_{0}^{\infty} e^{- {[x]}^{2}} = \sum_{n ⩾ 0} e^{- n^{2}} .

Answered by alex041103 last updated on 22/Feb/18

\int_{0}^{\infty} e^{- {[x]}^{2}} d x = \underset{n = 0}{\sum^{\infty}} \underset{n}{\int^{n + 1}} e^{- {[x]}^{2}} d x =

= \underset{n = 0}{\sum^{\infty}} \underset{n}{\int^{n + 1}} e^{- n^{2}} d x =

= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} \underset{n}{\int^{n + 1}} d x =

= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} x ∣_{n}^{n + 1} =

= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} (n + 1 - n) = \sum_{n ⩾ 0} e^{- n^{2}}