1) calculate ∫_0 ^1 ln(1+ix)dx and ∫_0 ^1 ln(1−ix)dx 2) find the value of ∫


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Question Number 49953 by maxmathsup by imad last updated on 12/Dec/18

$1) c a l c u l a t e \int_{0}^{1} l n (1 + i x) d x a n d \int_{0}^{1} l n (1 - i x) d x$ $2) f i n d t h e v a l u e o f \int_{0}^{1} l n (1 + x^{2}) d x .$
Commented by Abdo msup. last updated on 14/Dec/18

$1) w e h a v e 1 + i x = \sqrt{1 + x^{2}} (\frac{1}{\sqrt{1 + x^{2}}} + i \frac{x}{\sqrt{1 + x^{2}}}) = \sqrt{1 + x^{2}} e^{i a r c t a n (x)}$ $\Rightarrow \int_{0}^{1} l n (1 + i x) d x = \frac{1}{2} \int_{0}^{1} l n (1 + x^{2}) d x + i \int_{0}^{1} a r c t a n x d x b u t$ $\int_{0}^{1} l n (1 + x^{2}) d x =_{b y p a r t s} {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{2}} d x$ $= l n (2) - 2 \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} d x = l n (2) - 2 + 2 \int_{0}^{1} \frac{d x}{1 + x^{2}}$ $= l n (2) - 2 + 2 . \frac{π}{4} = l n (2) - 2 + \frac{π}{2} a l s o b y p a r t s$ $\int_{0}^{1} a r c t a n (x) d x = {[x a r c t a n x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x$ $= \frac{π}{4} - {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{1} = \frac{π}{4} - \frac{1}{2} l n (2) \Rightarrow$ $\int_{0}^{1} l n (1 + i x) d x = \frac{l n (2)}{2} - 1 + \frac{π}{4} + i (\frac{π}{4} - \frac{1}{2} l n (2))$
Commented by Abdo msup. last updated on 14/Dec/18

$a l s o 1 - i x = c o n j (1 + i x) = \sqrt{1 + x^{2}} e^{- i a r c t a n (x)} \Rightarrow$ $l n (1 - i x) = \frac{1}{2} l n (1 + x^{2}) - i a r c t a n x \Rightarrow$ $\int_{0}^{1} l n (1 - i x) d x = \frac{1}{2} \int_{0}^{1} l n (1 + x^{2}) d x - i \int_{0}^{1} a r c t a n x d x$ $= \frac{l n (2)}{2} - 1 + \frac{π}{4} - i (\frac{π}{4} - \frac{l n (2)}{2}) a n d w e s e e t h a t$ $\int_{0}^{1} l n (1 - i x) d x = c o n j (\int_{0}^{1} l n (1 + i x) d x) .$
Commented by Abdo msup. last updated on 14/Dec/18

$2) b y p a r t s \int_{0}^{1} l n (1 + x^{2}) d x = {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{2}} d x$ $= l n (2) - 2 \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} d x = l n (2) - 2 + 2 \int_{0}^{1} \frac{d x}{1 + x^{2}}$ $= l n (2) - 2 + 2 . \frac{π}{4} = l n (2) - 2 + \frac{π}{2}$ $c o m p l e x m e t h o d$ $\int_{0}^{1} l n (1 + x^{2}) d x = \int_{0}^{1} l n (1 + i x) (1 - i x)) d x$ $= \int_{0}^{1} l n (1 + i x) d x + \int_{0}^{1} l n (1 - i x) d x$ $= \frac{l n (2)}{2} - 1 + \frac{π}{4} + \frac{l n (2)}{2} - 1 + \frac{π}{4} = l n (2) - 2 + \frac{π}{2} .$
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