find-f-x-0-1-ln-x-e-t-dt-with-x-gt-0-

Question Number 67021 by mathmax by abdo last updated on 21/Aug/19

f i n d f (x) = \int_{0}^{1} l n (x + e^{- t}) d t w i t h x > 0

Commented by mathmax by abdo last updated on 25/Aug/19

w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{d t}{x + e^{- t}} d t = \frac{1}{x} \int_{0}^{1} \frac{d t}{1 + \frac{e^{- t}}{x}}

i f x > 1 w e h a v e f^{^{'}} (x) = \frac{1}{x} \int_{0}^{1} \sum_{n = 0}^{\infty} {(- 1)}^{n} {(\frac{e^{- t}}{x})}^{n}

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

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

f (x) = l n (x) - \sum_{n = 1}^{\infty} \frac{{(- e)}^{n}}{n} \int_{1}^{x} \frac{d t}{t^{n + 1}} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{1}^{x} \frac{d t}{t^{n + 1}} + C

\int_{1}^{x} \frac{d t}{t^{n + 1}} = \int_{1}^{x} t^{- n - 1} d t = {[\frac{1}{- n} t^{- n}]}_{1}^{x} = \frac{1}{n} (1 - \frac{1}{x^{n}}) \Rightarrow

f (x) = l n (x) - \sum_{n = 1}^{\infty} \frac{{(- e)}^{n}}{n^{2}} (1 - \frac{1}{x^{n}}) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} (1 - \frac{1}{x^{n}}) + C

C = f (1) = \int_{0}^{1} l n (1 + e^{- t}) d t \Rightarrow

f (x) = l n (x) - \sum_{n = 1}^{\infty} \frac{{(- e)}^{n}}{n^{2}} + \sum_{n = 1}^{\infty} \frac{{(- e)}^{n}}{n^{2} x^{n}} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}

- \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} x^{n}} \dots . b e c o n t i n u e d \dots .