calculate-n-0-1-n-4n-1-

Question Number 32025 by abdo imad last updated on 18/Mar/18

c a l c u l a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} .

Commented by abdo imad last updated on 20/Mar/18

l e t p u t S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} a n d S (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{4 n + 1}}{4 n + 1}

t h e r a d i u s o f S (x) i s r = 1 a n d f o r ∣ x ∣ ⩽ 1

S^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n} = \sum_{n = 0}^{\infty} {(- x^{4})}^{n} = \frac{1}{1 + x^{4}} \Rightarrow

S (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} + λ b u t λ = S (0) = 0 \Rightarrow S (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}}

l e t d e c o m p o s e f (t) = \frac{1}{1 + t^{4}}

f (t) = \frac{1}{{(1 + t^{2})}^{2} - 2 t^{2}} = \frac{1}{(t^{2} + 1 - \sqrt{2} t) (t^{2} + 1 + \sqrt{2} t)}

= \frac{a t + b}{(t^{2} - \sqrt{2} t + 1)} + \frac{c t + d}{t^{2} + \sqrt{2} t + 1}^{'} f (- t) = f (t) \Rightarrow

\frac{- a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- c t + d}{t^{2} - \sqrt{2} t + 1} = f (t) \Rightarrow c = - a a n d b = d \Rightarrow

f (t) = \frac{a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} + \sqrt{2} t + 1}

f (0) = 1 = 2 b \Rightarrow b = \frac{1}{2}

f (1) = \frac{1}{2} = \frac{a + b}{2 - \sqrt{2}} + \frac{b - a}{2 + \sqrt{2}} \Leftrightarrow (2 + \sqrt{2}) (a + b) + (2 - \sqrt{2}) (b - a) = 1

\Rightarrow 2 a + 2 b + \sqrt{2} a + \sqrt{2} b + 2 b - 2 a - \sqrt{2} b + \sqrt{2} a = 1 \Rightarrow

2 \sqrt{2} a + 4 b = 1 \Rightarrow 2 \sqrt{2} a + 2 = 1 \Rightarrow 2 \sqrt{2} a = - 1 \Rightarrow a = \frac{- 1}{2 \sqrt{2}} .

f (t) = \frac{\frac{- 1}{2 \sqrt{2}} t + \frac{1}{2}}{t^{2} - \sqrt{2} t + 1} + \frac{\frac{1}{2 \sqrt{2}} t + \frac{1}{2}}{t^{2} + \sqrt{2} t + 1}

f (t) = \frac{1}{2 \sqrt{2}} \frac{- t + \sqrt{2}}{t^{2} - \sqrt{2} t + 1} + \frac{1}{2 \sqrt{2}} \frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1}

\int_{0}^{x} \frac{d t}{1 + t^{4}} = \frac{1}{2 \sqrt{2}} \int_{0}^{x} \frac{- t + \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t + \frac{1}{2 \sqrt{2}} \int_{0}^{x} \frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t

\int_{0}^{x} \frac{- t + \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t = - \frac{1}{2} \int_{0}^{x} \frac{2 t - 2 \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t

= - \frac{1}{2} \int_{0}^{x} \frac{2 t - \sqrt{2}}{t^{2} - \sqrt{2} t + 1} + \frac{\sqrt{2}}{2} \int_{0}^{x} \frac{d t}{t^{2} - \sqrt{2} t + 1}

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

= l n (x^{2} - \sqrt{2} x + 1)

\int_{0}^{x} \frac{d t}{t^{2} - \sqrt{2} t + 1} = \int_{0}^{x} \frac{d t}{t^{2} - 2 \frac{\sqrt{2}}{2} t + \frac{1}{2} + \frac{3}{2}}

= \int_{0}^{x} \frac{d t}{{(t - \frac{\sqrt{2}}{2})}^{2} + \frac{3}{4}} =_{t - \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{2} u} \int_{- \frac{\sqrt{2}}{\sqrt{3}}}^{\frac{2}{\sqrt{3}} (x - \frac{\sqrt{2}}{2})} \frac{1}{\frac{3}{4} (1 + u^{2})} \frac{\sqrt{3}}{2} d u

= \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{\sqrt{2}}{\sqrt{3}}}^{\frac{2 x - \sqrt{2}}{\sqrt{3}}} \frac{d u}{1 + u^{2}} = \frac{2 \sqrt{3}}{3} (a r c t a n (\frac{2 x - \sqrt{2}}{\sqrt{3}}) - a r c t a n (\frac{\sqrt{2}}{\sqrt{3}}))

\int_{0}^{x} \frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t = \frac{1}{2} \int_{0}^{x} \frac{2 t + 2 \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t

= \frac{1}{2} \int_{0}^{x} \frac{2 t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t + \frac{\sqrt{2}}{2} \int_{0}^{x} \frac{d t}{t^{2} + \sqrt{2} t + 1} b u t

\int_{0}^{x} \frac{2 t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t = {[l n (t^{2} + \sqrt{2} t + 1)]}_{0}^{x} = l n (x^{2} + \sqrt{2} x + 1)

\int_{0}^{x} \frac{d t}{t^{2} + \sqrt{2} t + 1} = \frac{2 \sqrt{3}}{3} (a r c t a n (\frac{2 x + \sqrt{2}}{\sqrt{3}}) - a r c t a n (\frac{\sqrt{2}}{\sqrt{3}}))

2 \sqrt{2} \int_{0}^{x} \frac{d t}{1 + t^{4}} = - \frac{1}{2} l n (x^{2} - \sqrt{2} x + 1) + \frac{\sqrt{6}}{3} (a r c t a n (\frac{2 x - \sqrt{2}}{\sqrt{3}}) - a r c t a n (\frac{\sqrt{2}}{\sqrt{3}}))

+ \frac{1}{2} l n (x^{2} + \sqrt{2} x + 1) + \frac{\sqrt{6}}{3} (a r c t a n (\frac{2 x + \sqrt{2}}{\sqrt{3}}) - a r c t a n (\frac{\sqrt{2}}{\sqrt{3}}))

Commented by abdo imad last updated on 20/Mar/18

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