let A(x)= ∫_0 ^1 ln(1+ix^2 )dx find a simple form of f(x) (x∈R)


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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
$we have 1+ix^2 =(√(1+x^4 )) ((1/(√(1+x^4 ))) +i(x^2 /(√(1+x^4 )))) =r e^(iθ) ⇒r= (√(1+x^4 )) and cosθ =(1/(√(1+x^4 ))) ,sinθ=(x^2 /(√(1+x^2 ))) ⇒tanθ =x^2 ⇒θ =arctan(x^2 ) ⇒ 1+ix^2 =(√(1+x^4 )) e^(iarctan(x^2 )) ⇒ ln(1+ix^2 ) = (1/2)ln(1+x^4 ) +i arctan(x^2 ) ⇒ A(x) = (1/2)∫_0 ^1 ln(1+x^4 )dx +i ∫_0 ^1 arctan(x^2 )dx A +iB by parts B = [x arctan(x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 )/(1+x^4 ))dx B = (π/4) − ∫_0 ^1 ((2x^2 )/(1+x^4 ))dx is calculated after decomposition of F(x)= ((2x^2 )/(1+x^4 )) .... ln^′ (1+t) =(1/(1+t)) =Σ_(n=0) ^∞ (−1)^n t^n ⇒ ln(1+t)= Σ_(n=0) ^∞ (((−1)^n t^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/n) ⇒ln(1+x^4 ) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) ⇒ A =(1/2) ∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) =(1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) (1/(4n+1)) (A/2) = Σ_(n=1) ^∞ (((−1)^(n−1) )/(4n(4n+1))) = Σ_(n=1) ^∞ (−1)^(n−1) {(1/(4n)) −(1/(4n+1))} = (1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −Σ_(n=1) ^∞ (((−1)^(n−1) )/(4n+1)) =(1/4)ln(2) + Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) let w(x) = Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) x^(4n+1) w^′ (x) = Σ_(n=1) ^∞ (−1)^n x^(4n) = Σ_(n=1) ^∞ (−x^4 )^n = (1/(1+x^4 )) ⇒w(x)= ∫_0 ^x (dt/(1+t^4 )) + c but c=w(0)=0 ⇒ Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) = ∫_0 ^1 (dt/(1+t^4 )) ...the value of this integral is known after decoposition... =$
$w e h a v e 1 + {i x}^{2} = \sqrt{1 + x^{4}} (\frac{1}{\sqrt{1 + x^{4}}} + i \frac{x^{2}}{\sqrt{1 + x^{4}}})$ $= r e^{i θ} \Rightarrow r = \sqrt{1 + x^{4}} a n d c o s θ = \frac{1}{\sqrt{1 + x^{4}}}, s i n θ = \frac{x^{2}}{\sqrt{1 + x^{2}}}$ $\Rightarrow t a n θ = x^{2} \Rightarrow θ = a r c t a n (x^{2}) \Rightarrow$ $1 + {i x}^{2} = \sqrt{1 + x^{4}} e^{i a r c t a n (x^{2})} \Rightarrow$ $l n (1 + {i x}^{2}) = \frac{1}{2} l n (1 + x^{4}) + i a r c t a n (x^{2}) \Rightarrow$ $A (x) = \frac{1}{2} \int_{0}^{1} l n (1 + x^{4}) d x + i \int_{0}^{1} a r c t a n (x^{2}) d x$ $A + i B$ $b y p a r t s B = {[x a r c t a n (x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{4}} d x$ $B = \frac{π}{4} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{4}} d x i s c a l c u l a t e d a f t e r$ $d e c o m p o s i t i o n o f F (x) = \frac{2 x^{2}}{1 + x^{4}} . . . .$ ${l n}^{^{'}} (1 + t) = \frac{1}{1 + t} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n} \Rightarrow$ $l n (1 + t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} t^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{n}}{n}$ $\Rightarrow l n (1 + x^{4}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{4 n} \Rightarrow$ $A = \frac{1}{2} \int_{0}^{1} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{4 n}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \frac{1}{4 n + 1}$ $\frac{A}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{4 n (4 n + 1)} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{4 n} - \frac{1}{4 n + 1}}$ $= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{4 n + 1}$ $= \frac{1}{4} l n (2) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1}$ $l e t w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 n + 1}$ $w^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{4 n} = \sum_{n = 1}^{\infty} {(- x^{4})}^{n}$ $= \frac{1}{1 + x^{4}} \Rightarrow w (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} + c b u t c = w (0) = 0$ $\Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \int_{0}^{1} \frac{d t}{1 + t^{4}} . . . t h e v a l u e o f t h i s$ $i n t e g r a l i s k n o w n a f t e r d e c o p o s i t i o n . . .$ $=$
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