let-A-x-0-1-ln-1-ix-2-dx-find-a-simple-form-of-f-x-x-R-

Question Number 34771 by abdo mathsup 649 cc last updated on 10/May/18

l e t A (x) = \int_{0}^{1} l n (1 + {i x}^{2}) d x

f i n d a s i m p l e f o r m o f f (x) (x \in R)

Commented by abdo mathsup 649 cc last updated on 13/May/18

w e h a v e 1 + {i x}^{2} = \sqrt{1 + x^{4}} (\frac{1}{\sqrt{1 + x^{4}}} + i \frac{x^{2}}{\sqrt{1 + x^{4}}})

= r e^{i θ} \Rightarrow r = \sqrt{1 + x^{4}} a n d c o s θ = \frac{1}{\sqrt{1 + x^{4}}}, s i n θ = \frac{x^{2}}{\sqrt{1 + x^{2}}}

\Rightarrow t a n θ = x^{2} \Rightarrow θ = a r c t a n (x^{2}) \Rightarrow

1 + {i x}^{2} = \sqrt{1 + x^{4}} e^{i a r c t a n (x^{2})} \Rightarrow

l n (1 + {i x}^{2}) = \frac{1}{2} l n (1 + x^{4}) + i a r c t a n (x^{2}) \Rightarrow

A (x) = \frac{1}{2} \int_{0}^{1} l n (1 + x^{4}) d x + i \int_{0}^{1} a r c t a n (x^{2}) d x

A + i B

b y p a r t s B = {[x a r c t a n (x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{4}} d x

B = \frac{π}{4} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{4}} d x i s c a l c u l a t e d a f t e r

d e c o m p o s i t i o n o f F (x) = \frac{2 x^{2}}{1 + x^{4}} \dots .

{l n}^{^{'}} (1 + t) = \frac{1}{1 + t} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n} \Rightarrow

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

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

A = \frac{1}{2} \int_{0}^{1} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{4 n}

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

\frac{A}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{4 n (4 n + 1)} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{4 n} - \frac{1}{4 n + 1}}

= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{4 n + 1}

= \frac{1}{4} l n (2) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1}

l e t w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 n + 1}

w^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{4 n} = \sum_{n = 1}^{\infty} {(- x^{4})}^{n}

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

\Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \int_{0}^{1} \frac{d t}{1 + t^{4}} \dots t h e v a l u e o f t h i s

i n t e g r a l i s k n o w n a f t e r d e c o p o s i t i o n \dots

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