let f(x) =∫_0 ^1 ln(1+ixt)dt calculate f^, (x) (x from R).


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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
$we have 1+ixt =(√(1+x^2 t^2 ))((1/(√(1+x^2 t^2 ))) +i((xt)/(√(1+x^2 t^2 )))) =r e^(iθ) ⇒r=(√(1+x^2 t^2 )) and cosθ =(1/(√(1+x^2 t^2 ))) sinθ =((xt)/(√(1+x^2 t^2 ))) ⇒tanθ =xt ⇒θ=arctan(xt)⇒ ln(1+xt) =ln(r)+iθ =(1/2)ln(1+x^2 t^2 )+i arctan(xt) ⇒f(x)=(1/2)∫_0 ^1 ln(1+x^2 t^2 )dt +i ∫_0 ^1 arctan(xt)dt⇒ f^′ (x)=(1/2) ∫_0 ^1 ((2xt^2 )/(1+x^2 t^2 ))dt +i ∫_0 ^1 (t/(1+x^2 t^2 ))dt . = ∫_0 ^1 ((xt^2 )/(1+x^2 t^2 )) dt +i ∫_0 ^1 (t/(1+x^2 t^2 ))dt but for x≠0 ∫_0 ^1 ((xt^2 )/(1+x^2 t^2 )) dt =(1/x) ∫_0 ^1 ((x^2 t^2 +1−1)/(1+x^2 t^2 ))dt =(1/x) −(1/x) ∫_0 ^1 (dt/(1+x^2 t^2 )) =_(xt =u) (1/x) −(1/x)∫_0 ^x (1/(1+u^2 ))(du/x) =(1/x) −(1/x^2 ) arctan(x) also ∫_0 ^1 (t/(1+x^2 t^2 )) dt =_(xt =u) ∫_0 ^x (u/(x(1+u^2 ))) (du/x) =(1/x^2 ) ∫_0 ^x (u/(1+u^2 ))du =(1/(2x^2 ))ln∣1+x^2 ∣ ⇒ f^′ (x)=(1/x) −((arctanx)/x^2 ) +i{(1/(2x^2 ))ln(1+x^2 } .$
$w e h a v e 1 + i x t = \sqrt{1 + x^{2} t^{2}} (\frac{1}{\sqrt{1 + x^{2} t^{2}}} + i \frac{x t}{\sqrt{1 + x^{2} t^{2}}})$ $= r e^{i θ} \Rightarrow r = \sqrt{1 + x^{2} t^{2}} a n d c o s θ = \frac{1}{\sqrt{1 + x^{2} t^{2}}}$ $s i n θ = \frac{x t}{\sqrt{1 + x^{2} t^{2}}} \Rightarrow t a n θ = x t \Rightarrow θ = a r c t a n (x t) \Rightarrow$ $l n (1 + x t) = l n (r) + i θ = \frac{1}{2} l n (1 + x^{2} t^{2}) + i a r c t a n (x t)$ $\Rightarrow f (x) = \frac{1}{2} \int_{0}^{1} l n (1 + x^{2} t^{2}) d t + i \int_{0}^{1} a r c t a n (x t) d t \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2} \int_{0}^{1} \frac{2 {x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t .$ $= \int_{0}^{1} \frac{{x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t b u t f o r x \neq 0$ $\int_{0}^{1} \frac{{x t}^{2}}{1 + x^{2} t^{2}} d t = \frac{1}{x} \int_{0}^{1} \frac{x^{2} t^{2} + 1 - 1}{1 + x^{2} t^{2}} d t$ $= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 + x^{2} t^{2}} =_{x t = u} \frac{1}{x} - \frac{1}{x} \int_{0}^{x} \frac{1}{1 + u^{2}} \frac{d u}{x}$ $= \frac{1}{x} - \frac{1}{x^{2}} a r c t a n (x) a l s o$ $\int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t =_{x t = u} \int_{0}^{x} \frac{u}{x (1 + u^{2})} \frac{d u}{x}$ $= \frac{1}{x^{2}} \int_{0}^{x} \frac{u}{1 + u^{2}} d u = \frac{1}{2 x^{2}} l n ∣ 1 + x^{2} ∣ \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x} - \frac{a r c t a n x}{x^{2}} + i {\frac{1}{2 x^{2}} l n (1 + x^{2}} .$
Commented by prof Abdo imad last updated on 29/Aug/18

$a n o t h e r m e t h o d b y c o m p l e x d e r o v a t i o n$ $f^{^{'}} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (l n (1 + i x t)) d t$ $= \int_{0}^{1} \frac{i t}{1 + i x t} d t = \int_{0}^{1} \frac{i t (1 - i x t)}{1 + x^{2} t^{2}} d t$ $= \int_{0}^{1} \frac{{x t}^{2}}{1 + x^{2} t^{2}} d t + i \int_{0}^{1} \frac{t d t}{1 + x^{2} t^{2}} = . . . .$
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