let-f-x-0-1-ln-1-ixt-dt-calculate-f-x-x-from-R-

Question Number 42603 by maxmathsup by imad last updated on 28/Aug/18

l e t f (x) = \int_{0}^{1} l n (1 + i x t) d t c a l c u l a t e f^{,} (x) (x f r o m R) .

Commented by prof Abdo imad last updated on 29/Aug/18

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

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

s i n θ = \frac{x t}{\sqrt{1 + x^{2} t^{2}}} \Rightarrow t a n θ = x t \Rightarrow θ = a r c t a n (x t) \Rightarrow

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

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

f^{^{'}} (x) = \frac{1}{2} \int_{0}^{1} \frac{2 {x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t .

= \int_{0}^{1} \frac{{x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t b u t f o r x \neq 0

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

= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 + x^{2} t^{2}} =_{x t = u} \frac{1}{x} - \frac{1}{x} \int_{0}^{x} \frac{1}{1 + u^{2}} \frac{d u}{x}

= \frac{1}{x} - \frac{1}{x^{2}} a r c t a n (x) a l s o

\int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t =_{x t = u} \int_{0}^{x} \frac{u}{x (1 + u^{2})} \frac{d u}{x}

= \frac{1}{x^{2}} \int_{0}^{x} \frac{u}{1 + u^{2}} d u = \frac{1}{2 x^{2}} l n ∣ 1 + x^{2} ∣ \Rightarrow

f^{^{'}} (x) = \frac{1}{x} - \frac{a r c t a n x}{x^{2}} + i {\frac{1}{2 x^{2}} l n (1 + x^{2}} .

Commented by prof Abdo imad last updated on 29/Aug/18

a n o t h e r m e t h o d b y c o m p l e x d e r o v a t i o n

f^{^{'}} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (l n (1 + i x t)) d t

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

= \int_{0}^{1} \frac{{x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t d t}{1 + x^{2} t^{2}} = \dots .