Find-1-1-E-x-1-x-dx-

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

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) + \dots 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)