let-give-x-gt-0-1-prove-that-0-1-dt-1-t-x-n-0-1-n-nx-1-2-find-n-0-1-n-n-1-and-n-0-1-n-2n-1-3-find-n-1-1-n-

Question Number 29978 by abdo imad last updated on 14/Feb/18

l e t g i v e x > 0

1) p r o v e t h a t \int_{0}^{1} \frac{d t}{1 + t^{x}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n x + 1}

2) f i n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}

3) f i n d \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} .

Commented by abdo imad last updated on 16/Feb/18

1) f o r t \in] 0, 1] t^{x} = e^{x l n t} < 1 \Rightarrow \int_{0}^{1} \frac{d t}{1 + t^{x}} = \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n x}) d t

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

2) w e h a v e p r o v e d t h a t A (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n x + 1} = \int_{0}^{1} \frac{d t}{1 + t^{x}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = A (1) = \int_{0}^{1} \frac{d t}{1 + t} = {[l n (1 + t)]}_{0}^{1} = l n (2)

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} = A (2) = \int_{0}^{1} \frac{d t}{1 + t^{2}} = {[a r c t a n t]}_{0}^{1} = \frac{π}{4}

3) w e h a v e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} = A (3) = \int_{0}^{1} \frac{d t}{1 + t^{3}}

w e h a v e \int_{0}^{\infty} \frac{d t}{1 + t^{3}} = \int_{0}^{1} \frac{d t}{1 + t^{3}} + \int_{1}^{+ \infty} \frac{d t}{1 + t^{3}} t h e c h . t = \frac{1}{u} g i v e

\int_{1}^{\infty} \frac{d t}{1 + t^{3}} = \int_{0}^{1} \frac{1}{1 + \frac{1}{u^{3}}} \frac{d u}{u^{2}} = \int_{0}^{1} \frac{d u}{u^{2} + \frac{1}{u}} = \int_{0}^{1} \frac{u d u}{1 + u^{2}}

= \frac{1}{2} {[l n (1 + u^{2})]}_{0}^{1} = \frac{1}{2} l n 2 t h e c h . t^{3} = u g i v e

\int_{0}^{\infty} \frac{d t}{1 + t^{3}} = \int_{0}^{\infty} \frac{1}{1 + u} \frac{1}{3} u^{\frac{1}{3} - 1} d u = \frac{1}{3} \int_{0}^{\infty} \frac{u^{\frac{1}{3} - 1}}{1 + u} d u

= \frac{1}{3} \frac{π}{s i n (\frac{π}{3})} = \frac{π}{3} \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2 π}{3 \sqrt{3}} \Rightarrow

\int_{0}^{1} \frac{d t}{1 + t^{3}} = \int_{0}^{\infty} \frac{d t}{1 + t^{3}} - \int_{1}^{+ \infty} \frac{d t}{1 + t^{3}} = \frac{2 π}{3 \sqrt{3}} - \frac{1}{2} l n (2) \Rightarrow

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} = \frac{2 π}{3 \sqrt{3}} - \frac{1}{2} l n (2) .

Commented by abdo imad last updated on 16/Feb/18

f o r Q 3) w e h a v e u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)}

w i t h 0 < a < 1 .