calculate ∫_0 ^∞ (e^(−[x]) /(x+1))dx


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Question Number 87013 by mathmax by abdo last updated on 01/Apr/20

$c a l c u l a t e \int_{0}^{\infty} \frac{e^{- [x]}}{x + 1} d x$
Commented by mathmax by abdo last updated on 02/Apr/20
$I =∫_1 ^∞ (e^(−[x]) /(x+1))dx ⇒ I =Σ_(n=1) ^∞ ∫_n ^(n+1) (e^(−n) /(x+1))dx =Σ_(n=1) ^∞ e^(−n) [ln(x+1)]_n ^(n+1) =Σ_(n=1) ^∞ e^(−n) {ln(n+1)−ln(n)} =Σ_(n=1) ^∞ e^(−n) ln(((n+1)/n)) =Σ_(n=1) ^∞ e^(−n) ln(1+(1/n)) and this serie is convergent.$
$I = \int_{1}^{\infty} \frac{e^{- [x]}}{x + 1} d x \Rightarrow I = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{e^{- n}}{x + 1} d x$ $= \sum_{n = 1}^{\infty} e^{- n} {[l n (x + 1)]}_{n}^{n + 1} = \sum_{n = 1}^{\infty} e^{- n} {l n (n + 1) - l n (n)}$ $= \sum_{n = 1}^{\infty} e^{- n} l n (\frac{n + 1}{n}) = \sum_{n = 1}^{\infty} e^{- n} l n (1 + \frac{1}{n}) a n d t h i s s e r i e i s$ $c o n v e r g e n t .$
Commented by mathmax by abdo last updated on 02/Apr/20

$f o r g i v e t h e Q i s c a l c u l a t e \int_{1}^{+ \infty} \frac{e^{- [x]}}{x + 1} d x$
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