0-1-1-1-x-1-x-dx-is-fractional-part-is-floor-function-

Question Number 95650 by M±th+et+s last updated on 26/May/20

\int_{0}^{1} {{(- 1)}^{⌊ \frac{1}{x} ⌋} \frac{1}{x}} d x

{. .} i s f r a c t i o n a l p a r t

⌊ . . ⌋ i s f l o o r f u n c t i o n

Answered by mathmax by abdo last updated on 26/May/20

A = \int_{0}^{1} {{(- 1)}^{[\frac{1}{x}]} \times \frac{1}{x}} dx we do the changement \frac{1}{x} = t \Rightarrow

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

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

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

E = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{{(- 1)}^{n}}{t} dt = \sum_{n = 1}^{\infty} {(- 1)}^{n} (\ln (n + 1) - \ln (n))

= \sum_{n = 1}^{\infty} {(- 1)}^{n} \ln (1 + \frac{1}{n}) (serie is convergent due to

{(- 1)}^{n} \ln (1 + \frac{1}{n}) \sim \frac{{(- 1)}^{n}}{n})

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

= \sum_{k = 1}^{\infty} \int_{2 k}^{2 k + 1} \frac{[t]}{t^{2}} dt + \sum_{k = 0}^{\infty} \int_{2 k + 1}^{2 k + 2} \frac{[- t]}{t^{2}} dt

we have \sum_{k = 1}^{\infty} \int_{2 k}^{2 k + 1} \frac{[t]}{t^{2}} dt = \sum_{k = 1}^{\infty} \int_{2 k}^{2 k + 1} \frac{2 k}{t^{2}} dt

= \sum_{k = 1}^{\infty} (2 k) {[- \frac{1}{t}]}_{2 k}^{2 k + 1} = 2 \sum_{k = 1}^{\infty} k (\frac{1}{2 k} - \frac{1}{2 k + 1})

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

2 k + 1 < t < 2 k + 2 \Rightarrow - 2 k - 2 < - t < - 2 k - 1 \Rightarrow

[- t] = - 2 k - 2 \Rightarrow \sum_{k = 0}^{\infty} \int_{2 k + 1}^{2 k + 2} \frac{[- t]}{t^{2}} dt

= \sum_{k = 0}^{\infty} - \int_{2 k + 1}^{2 k + 2} \frac{2 k + 2}{t^{2}} dt

= - \sum_{k = 0}^{\infty} (2 k + 2) {[- \frac{1}{t}]}_{2 k + 1}^{2 k + 2} = \sum_{k = 0}^{\infty} (2 k + 2) (\frac{1}{2 k + 2} - \frac{1}{2 k + 1})

= \sum_{k = 0}^{\infty} (1 - \frac{2 k + 2}{2 k + 1}) \dots . its seems that this integral is

divefgent \dots!