prove-that-0-1-1-1-t-a-2-dt-n-0-1-n-2-n-na-1-2-find-the-value-of-n-0-1-n-2-n-3n-1-

Question Number 38118 by maxmathsup by imad last updated on 21/Jun/18

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

2) f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n} (3 n + 1)}

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

w e h a v e ∣ \frac{t^{a}}{2} ∣< 1 \Rightarrow \int_{0}^{1} \frac{1}{1 + \frac{t^{a}}{2}} d t

= \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{t^{n a}}{2^{n}}) d t

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n}} \int_{0}^{1} t^{n a} d t

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

2) w e h a v e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2^{n} (3 n + 1)} = \int_{0}^{1} \frac{d t}{1 + \frac{t^{3}}{2}}

= 2 \int_{0}^{1} \frac{d t}{2 + t^{3}} l e t α / α^{3} = 2 \Rightarrow α =^{3} \sqrt{2}

\int_{0}^{1} \frac{d t}{t^{3} + 2} = \int_{0}^{1} \frac{d t}{t^{3} + α^{3}} l e t d d c o m p o s e

F (t) = \frac{1}{t^{3} + α^{3}}

F (t) = \frac{1}{(t + α) (t^{2} - α t + α^{2})} = \frac{a}{t + α} + \frac{b t + c}{t^{2} - α t + α^{2}}

\dots b e c o n t i n u e d \dots