prove that ∫_0 ^∞ e^(−[x]^2 ) = Σ


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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

	$\int_{0}^{\infty} e^{- {[x]}^{2}} d x = \underset{n = 0}{\sum^{\infty}} \underset{n}{\int^{n + 1}} e^{- {[x]}^{2}} d x =$ $= \underset{n = 0}{\sum^{\infty}} \underset{n}{\int^{n + 1}} e^{- n^{2}} d x =$ $= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} \underset{n}{\int^{n + 1}} d x =$ $= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} x ∣_{n}^{n + 1} =$ $= \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} (n + 1 - n) = \sum_{n ⩾ 0} e^{- n^{2}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum