1-calculate-0-dx-1-e-nx-with-n-integr-natural-and-n-1-2-conclude-the-value-of-k-1-1-k-1-k-

Question Number 65383 by mathmax by abdo last updated on 29/Jul/19

1) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{1 + e^{n x}} w i t h n i n t e g r n a t u r a l a n d n ⩾ 1

2) c o n c l u d e t h e v a l u e o f \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k}

Commented by mathmax by abdo last updated on 30/Jul/19

1) l e t A_{n} = \int_{0}^{\infty} \frac{d x}{1 + e^{n x}} c h a n g e m e n t e^{n x} = t v i v e n x = l n t \Rightarrow

A_{n} = \int_{1}^{+ \infty} \frac{d t}{n t (1 + t)} = \frac{1}{n} \int_{1}^{+ \infty} {\frac{1}{t} - \frac{1}{t + 1}} d t = \frac{1}{n} {[l n ∣ \frac{t}{t + 1} ∣]}_{1}^{+ \infty}

= \frac{1}{n} {- l n (\frac{1}{2})} = \frac{l n (2)}{n}

2) w e h a v e A_{n} = \int_{0}^{\infty} \frac{d x}{1 + e^{n x}} = \int_{0}^{\infty} \frac{e^{- n x}}{1 + e^{- n x}} d x

= \int_{0}^{\infty} e^{- n x} {\sum_{k = 0}^{\infty} {(- 1)}^{k} e^{- k n x}} = \sum_{k = 0}^{\infty} {(- 1)}^{k} \int_{0}^{\infty} e^{- (n + k n) x} d x

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

= \frac{1}{n} \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k} b u t A_{n} = \frac{l n (2)}{n} \Rightarrow \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k} = l n (2) .