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Question Number 42603 by maxmathsup by imad last updated on 28/Aug/18
let f(x) =∫_0 ^1 ln(1+ixt)dt calculate f^, (x) (x from R).
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{ixt}\right){dt}\:\:{calculate}\:{f}^{,} \left({x}\right)\:\:\:\:\left({x}\:{from}\:{R}\right). \\ $$
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 } .
$${we}\:{have}\:\mathrm{1}+{ixt}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }}\:+{i}\frac{{xt}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:{and}\:{cos}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }} \\ $$$${sin}\theta\:=\frac{{xt}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }}\:\Rightarrow{tan}\theta\:={xt}\:\Rightarrow\theta={arctan}\left({xt}\right)\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{xt}\right)\:={ln}\left({r}\right)+{i}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)+{i}\:{arctan}\left({xt}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right){dt}\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({xt}\right){dt}\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{xt}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt}\:\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt}\:. \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xt}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:{dt}\:\:+{i}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt}\:{but}\:{for}\:{x}\neq\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xt}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:{dt}\:=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:=_{{xt}\:={u}} \:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\frac{{du}}{{x}} \\ $$$$=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{arctan}\left({x}\right)\:\:{also} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{t}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:{dt}\:=_{{xt}\:={u}} \:\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{{du}}{{x}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }{ln}\mid\mathrm{1}+{x}^{\mathrm{2}} \mid\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}}{{x}}\:−\frac{{arctanx}}{{x}^{\mathrm{2}} }\:\:+{i}\left\{\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right\}\:.\right. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 29/Aug/18
another method by complex derovation f^′ (x) =∫_0 ^1 (∂/∂x)(ln(1+ixt))dt = ∫_0 ^1 ((it)/(1+ixt))dt =∫_0 ^1 ((it(1−ixt))/(1+x^2 t^2 ))dt = ∫_0 ^1 ((xt^2 )/(1+x^2 t^2 ))dt +i ∫_0 ^1 ((tdt)/(1+x^2 t^2 )) =....
$${another}\:{method}\:{by}\:{complex}\:{derovation} \\ $$$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\partial}{\partial{x}}\left({ln}\left(\mathrm{1}+{ixt}\right)\right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{it}}{\mathrm{1}+{ixt}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{it}\left(\mathrm{1}−{ixt}\right)}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xt}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt}\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{tdt}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:=…. \\ $$

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