calculus-please-evaluate-0-1-ln-x-dx-

Question Number 126969 by mnjuly1970 last updated on 25/Dec/20

c a l c u l u s

p l e a s e e v a l u a t e ::

ϕ = \int_{0}^{1} ⌊ l n (x) ⌋ d x = ?

Answered by mindispower last updated on 25/Dec/20

l n (x) = t

= \int_{- \infty}^{0} [t] e^{t}

= \underset{k = 0}{\sum^{\infty}} \int_{- k - 1}^{- k} [t] e^{t} d t

= - \underset{k = 0}{\sum^{\infty}} \int_{- k - 1}^{- k} (k + 1) e^{t} =

= - \underset{k = 0}{\sum^{\infty}} (k + 1) (e^{- k} - e^{- k - 1}) d t

= - Σ ({k e}^{- k} - (k + 1) e^{- (k + 1)}) - \underset{k = 0}{\sum^{\infty}} e^{- k}

= - (\lim_{N \to \infty} (- (N + 1) e^{- (N + 1)}) + \frac{1 - e^{- (N + 1)}}{1 - e^{- 1}})

= - \frac{1}{1 - e^{- 1}} = - \frac{e}{e - 1}

Commented by mnjuly1970 last updated on 25/Dec/20

m e r c e y s i r m i d s p o w d r

g r a t e f u l \dots

Answered by mathmax by abdo last updated on 27/Dec/20

Φ = \int_{0}^{1} [lnx] dx we do the changement lnx = - t \Rightarrow x = e^{- t} \Rightarrow

Φ = - \int_{0}^{\infty} [- t] (- e^{- t}) dt = \int_{0}^{\infty} [- t] e^{- t} dt

= \sum_{n = 0}^{\infty} \int_{n}^{n + 1} [- t] e^{- t} dt we have n ⩽ t ⩽ n + 1 \Rightarrow - n - 1 ⩽ - t ⩽ - n \Rightarrow

[- t] = - (n + 1) \Rightarrow Φ = - \sum_{n = 0}^{\infty} \int_{n}^{n + 1} - (n + 1) e^{- t} dt

= \sum_{n = 0}^{\infty} (n + 1) {[- e^{- t}]}_{n}^{n + 1} = \sum_{n = 0}^{\infty} (n + 1) (e^{- n} - e^{- (n + 1)})

= \sum_{n = 0}^{\infty} (n + 1) e^{- n} - \sum_{n = 0}^{\infty} (n + 1) e^{- (n + 1)}

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

= (e - 1) \sum_{n = 1}^{\infty} n {(\frac{1}{e})}^{n} we have \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} for ∣ x ∣< 1 \Rightarrow

\sum_{n = 1}^{\infty} {nx}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} n {(\frac{1}{e})}^{n} = \frac{1}{e {(\frac{1}{e} - 1)}^{2}} = \frac{e}{{(1 - e)}^{2}} \Rightarrow Φ = (e - 1) \times \frac{e}{{(1 - e)}^{2}} = \frac{e}{1 - e}