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Question Number 39891 by math khazana by abdo last updated on 13/Jul/18

l e t f (x) = a r c t a n (2 x + 1)

1) c a l c u l a t e f^{(n)} (x)

2) c a l c u l a t e f^{(n)} (0)

3) d e v e l o p p f a t i n t e g r s e r i e

4) c a l c u l a t e \int_{0}^{1} f (x) d x

5) c a l c u l a t e \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{4 x^{2} + 4 x + 2} d x

Commented by maxmathsup by imad last updated on 13/Jul/18

1) w e h a v e f^{^{'}} (x) = \frac{2}{1 + {(2 x + 1)}^{2}} \Rightarrow f^{(n)} (x) = 2 {(\frac{1}{{(2 x + 1)}^{2} + 1})}^{(n - 1)}

l e t w (x) = \frac{1}{{(2 x + 1)}^{2} + 1} . l e t d e c o m p o s e w i n s i d e C (x)

w (x) = \frac{1}{(2 x + 1 - i) (2 x + 1 + i)} = \frac{1}{4 (x + \frac{1 - i}{2}) (x + \frac{1 + i}{2})}

= \frac{1}{4 (x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) (x + \frac{1}{\sqrt{2}} e^{i \frac{π}{4}})} = \frac{a}{x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} + \frac{b}{x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}}

a = {l i m}_{x \to - \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} (x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) w (x) = \frac{1}{4 \frac{1}{\sqrt{2}} (e^{\frac{i π}{4}} - e^{- \frac{i π}{4}})} = \frac{\sqrt{2}}{4 (2 i \frac{1}{\sqrt{2}})} = \frac{1}{4 i}

b = {l i m}_{x \to - \frac{1}{\sqrt{2}} e^{i \frac{π}{4}}} (x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}) w (x) = \frac{1}{4 \frac{1}{\sqrt{2}} (- e^{\frac{i π}{4}} + e^{- \frac{i π}{4}})} = \frac{\sqrt{2}}{- 4 (2 i \frac{1}{\sqrt{2}})}

= \frac{- 2}{8 i} = - \frac{1}{4 i} \Rightarrow w (x) = \frac{1}{4 i} {\frac{1}{x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} - \frac{1}{x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}}} \Rightarrow

w^{(n - 1)} = \frac{1}{4 i} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}} \Rightarrow

f^{(n)} (x) = 2 w^{(n - 1)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{1}{{(x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{1}{{(x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}} .

Commented by maxmathsup by imad last updated on 13/Jul/18

2) f o r x = 0 w e g e t

f^{(n)} (0) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{1}{{(\frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{1}{{(\frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}}

= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {{(\sqrt{2})}^{n} e^{\frac{i n π}{4}} - {(\sqrt{2})}^{n} e^{- \frac{i n π}{4}}}

= \frac{{(\sqrt{2})}^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} (2 i s i n (\frac{n π}{4})) \Rightarrow

f^{(n)} (0) = {(- 1)}^{n} (n - 1)! {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) .

Commented by maxmathsup by imad last updated on 14/Jul/18

3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

f (x) = \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} (n - 1)! {(\sqrt{2})}^{n}}{n!} x^{n}

= \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) x^{n} .

Commented by maxmathsup by imad last updated on 14/Jul/18

4) \int_{0}^{1} f (x) d x = \int_{0}^{1} a r c t a n (2 x + 1) d x b y p a r t s u^{^{'}} = 1 a n d v = a r c t a n (2 x + 1)

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

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

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

= \frac{1}{4} \int_{0}^{1} \frac{4 x + 2}{2 x^{2} + 2 x + 1} d x - \frac{1}{2} \int_{0}^{1} \frac{d x}{2 (x^{2} + x + \frac{1}{2})}

= \frac{1}{4} {[l n ∣ 2 x^{2} + 2 x + 1 ∣]}_{0}^{1} - \frac{1}{4} \int_{0}^{1} \frac{d x}{x^{2} + 2 \frac{1}{2} x + \frac{1}{4} + \frac{1}{4}}

= \frac{1}{4} l n (5) - \frac{1}{4} \int_{0}^{1} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{1}{4}} (c h . x + \frac{1}{2} = \frac{1}{2} t)

= \frac{1}{4} l n (5) - \frac{1}{4} \int_{0}^{1} \frac{1}{\frac{1}{4} (1 + t^{2})} \frac{d t}{2} = \frac{l n (5)}{4} - \frac{1}{2} \frac{π}{4} = \frac{l n (5)}{4} - \frac{π}{8} \Rightarrow

\int_{0}^{1} f (x) d x = a r c t a n (3) - \frac{l n (5)}{4} + \frac{π}{8} .

Commented by maxmathsup by imad last updated on 14/Jul/18

5) w e h a v e \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{4 x^{2} + 4 x + 2} d x = \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} a n d b y p a r t s

u^{^{'}} = \frac{1}{{(2 x + 1)}^{2} + 1} a n d v = a r c t a n (2 x + 1) w e g e t

\int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} d x = {[\frac{1}{2} a r c t a n (2 x + 1) . a r c t a n (2 x + 1)]}_{0}^{1}

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

= \frac{1}{2} {(a r c t a n (3))}^{2} - \frac{π^{2}}{32} - \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} \Rightarrow

\int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} d x = \frac{1}{4} {(a r c t a n (3))}^{2} - \frac{π^{2}}{64} .