0-1-1-1-1-x-dx-

Question Number 96514 by Farruxjano last updated on 02/Jun/20

\underset{0}{\int^{1}} \frac{1}{1 + [\frac{1}{x}]} d x = ?

Commented by Farruxjano last updated on 02/Jun/20

I {don}^{'} t know english perfect, but i can say

[x] - the whole part of x . x = [x] + {x}

Commented by prakash jain last updated on 02/Jun/20

[\frac{1}{x}] = ⌊ \frac{1}{x} ⌋ floor function ?

Commented by prakash jain last updated on 02/Jun/20

\int_{0}^{1} \frac{1}{1 + ⌊ \frac{1}{x} ⌋} =

x \in (\frac{1}{2}, 1] \Rightarrow ⌊ \frac{1}{x} ⌋ = 1 area = \frac{1}{1 + ⌊ \frac{1}{x} ⌋} \times (1 - \frac{1}{2})

x \in (\frac{1}{3}, \frac{1}{2}] \Rightarrow ⌊ \frac{1}{x} ⌋ = 2 area = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{3})

x \in (\frac{1}{n + 1}, \frac{1}{n}] \Rightarrow ⌊ \frac{1}{x} ⌋ = n

area = \frac{1}{n + 1} \times (\frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{n + 1}

= \frac{1}{n {(n + 1)}^{2}}

I = \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 1)}^{2}} = 2 - \frac{π^{2}}{6}

Answered by abdomathmax last updated on 02/Jun/20

I = \int_{0}^{1} \frac{dx}{1 + [\frac{1}{x}]} changement \frac{1}{x} = t give

I = - \int_{1}^{+ \infty} \frac{1}{1 + [t]} (- \frac{dt}{t^{2}}) = \int_{1}^{+ \infty} \frac{dt}{t^{2} (1 + [t])}

= \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{1}{n + 1} \times \frac{dt}{t^{2}} = \sum_{n = 1}^{\infty} \frac{1}{n + 1} {[- \frac{1}{t}]}_{n}^{n + 1}

= \sum_{n = 1}^{\infty} \frac{1}{n + 1} (\frac{1}{n} - \frac{1}{n + 1})

= \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} - \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}}

\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} = \lim_{n \to + \infty} \sum_{k = 1}^{n} (\frac{1}{k (k + 1)})

= \lim_{n \to + \infty} \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1})

= \lim_{n \to + \infty} (1 - \frac{1}{n + 1}) = 1

\sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}} = \sum_{n = 2}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} - 1 \Rightarrow

I = 1 - (\frac{π^{2}}{6} - 1) = 2 - \frac{π^{2}}{6}