1-calculate-0-1-ln-1-ix-dx-and-0-1-ln-1-ix-dx-2-find-the-value-of-0-1-ln-1-x-2-dx-

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} .