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Question Number 213945 by Spillover last updated on 22/Nov/24

	Answered by mathmax last updated on 23/Nov/24
	$I=∫_0 ^∞ e^(−[x](1+x−[x])) dx =1+Σ_(n=1) ^∞ ∫_n ^(n+1) e^(−n(1+x−n)) dx =1+Σ_(n=1) ^∞ ∫_n ^(n+1) e^(−n+n^2 ) e^(−nx) dx =1+Σ_(n=1) ^∞ e^(−n+n^2 ) [−(1/n)e^(−nx) ]_n ^(n+1) =1−Σ_(n=1) ^∞ (e^(n^2 −n) /n){ e^(−n^2 −n) −e^(−n^2 ) } =1−Σ_(n=1) ^∞ (e^(−2n) /n)+Σ_(n=1) ^∞ (e^(−n) /n) on rappelle que Σ (z^n /n)=−ln(1−z) pour ∣z∣<1 I=1+ln(1−e^(−2) )−ln(1−e^(−1) ) =1+ln(1−(1/e^2 ))−ln(1−(1/e))$
	$I = \int_{0}^{\infty} e^{- [x] (1 + x - [x])} d x$ $= 1 + \sum_{n = 1}^{\infty} \int_{n}^{n + 1} e^{- n (1 + x - n)} d x$ $= 1 + \sum_{n = 1}^{\infty} \int_{n}^{n + 1} e^{- n + n^{2}} e^{- n x} d x$ $= 1 + \sum_{n = 1}^{\infty} e^{- n + n^{2}} {[- \frac{1}{n} e^{- n x}]}_{n}^{n + 1}$ $= 1 - \sum_{n = 1}^{\infty} \frac{e^{n^{2} - n}}{n} {e^{- n^{2} - n} - e^{- n^{2}}}$ $= 1 - \sum_{n = 1}^{\infty} \frac{e^{- 2 n}}{n} + \sum_{n = 1}^{\infty} \frac{e^{- n}}{n}$ $o n r a p p e l l e q u e Σ \frac{z^{n}}{n} = - l n (1 - z) p o u r ∣ z ∣< 1$ $I = 1 + l n (1 - e^{- 2}) - l n (1 - e^{- 1})$ $= 1 + l n (1 - \frac{1}{e^{2}}) - l n (1 - \frac{1}{e})$
	Commented by mr W last updated on 23/Nov/24

	$= 1 + \ln (1 + \frac{1}{e})$
	Commented by mehdee7396 last updated on 23/Nov/24

	$= l n (e + 1)$
	Commented by Spillover last updated on 26/Nov/24

	$r i g h t$
	Answered by Spillover last updated on 23/Nov/24

	Answered by Spillover last updated on 23/Nov/24

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