Find ∫_1 ^∞ ((1/(E(x))) −(1/x))dx


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Question Number 66446 by ~ À ® @ 237 ~ last updated on 15/Aug/19

$F i n d \int_{1}^{\infty} (\frac{1}{E (x)} - \frac{1}{x}) d x$
Commented by mathmax by abdo last updated on 15/Aug/19
$let A =∫_1 ^(+∞) ((1/([x]))−(1/x))dx ⇒A =Σ_(n=1) ^∞ ∫_n ^(n+1) ((1/n)−(1/x))dx =Σ_(n=1) ^∞ ( (1/n)−∫_n ^(n+1) (dx/x)) =Σ_(n=1) ^∞ ((1/n)−ln(n+1)+ln(n)) let S_n =Σ_(k=1) ^n ((1/k)−ln(k+1)+ln(k)) we have A=lim_(n→+∞) S_n but S_(n ) =H_n +Σ_(k=1) ^n {ln(k)−ln(k+1)} =H_n +(ln(1)−ln(2)+ln(2)−ln(3)+...ln(n)−ln(n+1) =H_n −ln(n+1) =H_n −ln(n)−ln(1+(1/n)) but we know lim_(n→+∞) H_n −ln(n) =γ ⇒ A=lim_(n→+∞) S_n =γ (constant of Euler)$
$l e t A = \int_{1}^{+ \infty} (\frac{1}{[x]} - \frac{1}{x}) d x \Rightarrow A = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} (\frac{1}{n} - \frac{1}{x}) d x$ $= \sum_{n = 1}^{\infty} (\frac{1}{n} - \int_{n}^{n + 1} \frac{d x}{x}) = \sum_{n = 1}^{\infty} (\frac{1}{n} - l n (n + 1) + l n (n))$ $l e t S_{n} = \sum_{k = 1}^{n} (\frac{1}{k} - l n (k + 1) + l n (k)) w e h a v e A = {l i m}_{n \to + \infty} S_{n}$ $b u t S_{n} = H_{n} + \sum_{k = 1}^{n} {l n (k) - l n (k + 1)}$ $= H_{n} + (l n (1) - l n (2) + l n (2) - l n (3) + . . . l n (n) - l n (n + 1)$ $= H_{n} - l n (n + 1) = H_{n} - l n (n) - l n (1 + \frac{1}{n}) b u t w e k n o w$ ${l i m}_{n \to + \infty} H_{n} - l n (n) = γ \Rightarrow A = {l i m}_{n \to + \infty} S_{n} = γ$ $(c o n s t a n t o f E u l e r)$
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