find-the-value-of-I-0-ln-1-e-x-dx-nowing-that-n-1-1-n-2-pi-2-6-

Question Number 204569 by pticantor last updated on 21/Feb/24

f i n d t h e v a l u e o f

I = \int_{0}^{+ \infty} l n (1 + e^{- x}) d x n o w i n g t h a t

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6}

Answered by witcher3 last updated on 22/Feb/24

\forall x \in R_{+}, e^{- x} < 1

\ln (1 + e^{- x}) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{n + 1} e^{- (n + 1) x}

\int_{0}^{\infty} \ln (1 + e^{- x}) dx = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{n + 1} \int_{0}^{\infty} e^{- (n + 1) x} dx

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = η (2) = (1 - 2^{1 - 2}) ζ (2) = \frac{π^{2}}{12}

Commented by pticantor last updated on 23/Feb/24

p l s h o w d o y o u m a n a g e t o h a v e (1 - 2^{1 - 2}) ζ (2) ?

Commented by witcher3 last updated on 23/Feb/24

\sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = \sum_{n ⩾ 0} \frac{1}{{(2 n + 1)}^{2}} - \frac{1}{{4 n}^{2}}

= Σ (\frac{1}{n^{2}} - \frac{1}{{(2 n)}^{2}} - \frac{1}{{4 n}^{2}}) = Σ \frac{1}{n^{2}} - \frac{1}{2} Σ \frac{1}{n^{2}} = ζ (2) - \frac{1}{2} ζ (2) = (1 - 2^{1 - 2}) ζ (2)