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Question Number 34635 by abdo mathsup 649 cc last updated on 09/May/18

calculate A(α) = ∫_0 ^1 ln(1+αix)dx 2) calculate ∫_0 ^1 ln(1+ix) dx (i^2 =−1)

$${calculate}\:{A}\left(\alpha\right)\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+\alpha{ix}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{ix}\right)\:{dx}\:\:\:\:\left({i}^{\mathrm{2}} \:=−\mathrm{1}\right) \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/May/18

Commented by abdo mathsup 649 cc last updated on 10/May/18

we have 1+i(αx) =(√(1+α^2 x^2 )) ( (1/(√(1+α^2 x^2 ))) +i((αx)/(√(1+α^2 x^2 )))) =(√(1+α^2 x^2 )) e^(iθ) /cosθ =(1/(√(1+α^2 x^2 ))) and sinθ = ((αx)/(√(1+α^2 x^2 ))) ⇒ tanθ =αx ⇒θ =arctan(αx) ln(1 +iαx)=(1/2)ln(1+α^2 x^2 ) +i arctan(αx) ⇒ A(α) = (1/2) ∫_0 ^1 ln(1+α^2 x^2 )dx +i ∫_0 ^1 arctan(αx)dx =f(α) +i g(α) 2f^′ (α) = ∫_0 ^1 ((2αx^2 )/(1+α^2 x^2 ))dx = (2/α) ∫_0 ^1 ((1+α^2 x^2 −1)/(1+α^2 x^2 ))dx = (2/α) −(2/α) ∫_0 ^1 (dx/(1+α^2 x^2 )) =_(αx =u) (2/α) −(2/α) ∫_0 ^α (1/(1+u^2 )) (du/α) =(2/α) −(2/α^2 ) arctan(α) ⇒ f^′ (α) = (1/α) −((arctan(α))/α^2 ) ⇒ f(α)= ln∣α∣ − ∫^α ((arcrctant)/t^2 ) dt by parts ∫ ((arctant)/t^2 )dt = −(1/t) arctant −∫ ((−1)/t) (dt/(1+t^2 )) =−(1/t)arctant + ∫ (dt/(t(1+t^2 ))) =−((arctant)/t) + ∫ ((1/t) −(t/(1+t^2 )))dt =−((arctant)/t) +ln∣t∣ −(1/2) ln(1+t^2 ) ⇒ f(α)= ((arctan(α))/α) +(1/2)ln(1+α^2 ) +c c =lim_(α→.0) (f(α)−((arctan(α))/α) −(1/2)ln(1+α^2 ))=−1 f(α) =((arctan(α))/α) +(1/2)ln(1+α^2 ) −1 g(α) = ∫_0 ^1 arctan(αx)dx =_(αx =t) ∫_0 ^α arctant (dt/α) =(1/α) ∫_0 ^α arctant dt = (1/α){ [t arctant]_0 ^α −∫_0 ^α (t/(1+t^2 ))dt} = (1/α){ α arctan(α) −(1/2)ln(1+α^2 )} ⇒ A(α) = ((arctan(α))/α) +(1/2)ln(1+α^2 ) −1 i{ arctan(α) −(1/(2α))ln(1+α^2 )}

$${we}\:{have}\:\:\mathrm{1}+{i}\left(\alpha{x}\right)\:=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }\:\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }}\:+{i}\frac{\alpha{x}}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }}\right) \\ $$$$=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }\:\:{e}^{{i}\theta} \:\:\:/{cos}\theta\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }}\:{and} \\ $$$${sin}\theta\:=\:\frac{\alpha{x}}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }}\:\Rightarrow\:{tan}\theta\:=\alpha{x}\:\Rightarrow\theta\:={arctan}\left(\alpha{x}\right) \\ $$$${ln}\left(\mathrm{1}\:+{i}\alpha{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)\:+{i}\:{arctan}\left(\alpha{x}\right)\:\Rightarrow \\ $$$${A}\left(\alpha\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right){dx}\:\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\alpha{x}\right){dx} \\ $$$$={f}\left(\alpha\right)\:+{i}\:{g}\left(\alpha\right) \\ $$$$\mathrm{2}{f}^{'} \left(\alpha\right)\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}\alpha{x}^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{2}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=\:\frac{\mathrm{2}}{\alpha}\:\:−\frac{\mathrm{2}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$=_{\alpha{x}\:={u}} \:\:\frac{\mathrm{2}}{\alpha}\:\:−\frac{\mathrm{2}}{\alpha}\:\:\int_{\mathrm{0}} ^{\alpha} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\alpha} \\ $$$$=\frac{\mathrm{2}}{\alpha}\:\:−\frac{\mathrm{2}}{\alpha^{\mathrm{2}} }\:{arctan}\left(\alpha\right)\:\Rightarrow\:{f}^{'} \left(\alpha\right)\:=\:\frac{\mathrm{1}}{\alpha}\:−\frac{{arctan}\left(\alpha\right)}{\alpha^{\mathrm{2}} } \\ $$$$\Rightarrow\:{f}\left(\alpha\right)=\:{ln}\mid\alpha\mid\:−\:\:\int^{\alpha} \:\:\:\frac{{arcrctant}}{{t}^{\mathrm{2}} }\:{dt}\:\:{by}\:{parts} \\ $$$$\:\int\:\:\:\:\:\frac{{arctant}}{{t}^{\mathrm{2}} }{dt}\:=\:−\frac{\mathrm{1}}{{t}}\:{arctant}\:−\int\:\frac{−\mathrm{1}}{{t}}\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{{t}}{arctant}\:\:\:+\:\int\:\:\:\:\:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=−\frac{{arctant}}{{t}}\:+\:\int\:\:\left(\frac{\mathrm{1}}{{t}}\:−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$=−\frac{{arctant}}{{t}}\:\:+{ln}\mid{t}\mid\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)=\:\frac{{arctan}\left(\alpha\right)}{\alpha}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\:+{c} \\ $$$${c}\:={lim}_{\alpha\rightarrow.\mathrm{0}} \left({f}\left(\alpha\right)−\frac{{arctan}\left(\alpha\right)}{\alpha}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\right)=−\mathrm{1} \\ $$$${f}\left(\alpha\right)\:=\frac{{arctan}\left(\alpha\right)}{\alpha}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\:−\mathrm{1} \\ $$$${g}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\left(\alpha{x}\right){dx} \\ $$$$=_{\alpha{x}\:={t}} \:\:\int_{\mathrm{0}} ^{\alpha} \:\:\:\:{arctant}\:\frac{{dt}}{\alpha}\:\:=\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\alpha} \:\:{arctant}\:{dt} \\ $$$$=\:\frac{\mathrm{1}}{\alpha}\left\{\:\:\:\left[{t}\:{arctant}\right]_{\mathrm{0}} ^{\alpha} \:\:−\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right\} \\ $$$$=\:\frac{\mathrm{1}}{\alpha}\left\{\:\:\alpha\:{arctan}\left(\alpha\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\right\}\:\Rightarrow \\ $$$${A}\left(\alpha\right)\:=\:\frac{{arctan}\left(\alpha\right)}{\alpha}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\:−\mathrm{1} \\ $$$${i}\left\{\:{arctan}\left(\alpha\right)\:−\frac{\mathrm{1}}{\mathrm{2}\alpha}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\right\} \\ $$

Commented by abdo mathsup 649 cc last updated on 10/May/18

2) ∫_0 ^1 ln(1+ix)dx =A(1) =(π/4) +(1/2)ln(2) +i( (π/4) −(1/2) ln(2)) .

$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{ix}\right){dx}\:={A}\left(\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+{i}\left(\:\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right)\right)\:. \\ $$

Commented by abdo mathsup 649 cc last updated on 10/May/18

A(1) = (π/4) +(1/2)ln(2) −1 +i( (π/4) −(1/2) ln(2)) .

$${A}\left(\mathrm{1}\right)\:=\:\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:−\mathrm{1}\:+{i}\left(\:\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right)\right)\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

∫_0 ^1 ln(1+iαx)=∫_0 ^1 (1/2)ln(1^2 +α^2 x^2 )+itan^(−1) (((αx)/1))dx =I_1 +I_2 I_1 =(1/2)∫_0 ^1 (α^2 x^2 −((α^4 x^4 )/2)+((α^6 x^6 )/3)−((α^8 x^8 )/4)+.... dx using ln(1+x)=x−(x^2 /2)+(x^3 /3)− =(1/2)∣((α^2 x^3 )/3)−((α^4 x^5 )/(2×5))+((α^6 x^7 )/(3×7))−((α^8 x^9 )/(4×9)).....∣_0 ^1 =(1/2)((α^2 /3)−(α^4 /(10))+(α^6 /(21))−(α^8 /(36))...) I_2 =∫_0 ^1 itan^(−1) (αx) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{i}\alpha{x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}^{\mathrm{2}} +\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)+{itan}^{−\mathrm{1}} \left(\frac{\alpha{x}}{\mathrm{1}}\right){dx} \\ $$$$={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\alpha^{\mathrm{2}} {x}^{\mathrm{2}} −\frac{\alpha^{\mathrm{4}} {x}^{\mathrm{4}} }{\mathrm{2}}+\frac{\alpha^{\mathrm{6}} {x}^{\mathrm{6}} }{\mathrm{3}}−\frac{\alpha^{\mathrm{8}} {x}^{\mathrm{8}} }{\mathrm{4}}+....\:\:{dx}\right. \\ $$$${using}\:{ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}− \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{\alpha^{\mathrm{2}} {x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\alpha^{\mathrm{4}} {x}^{\mathrm{5}} }{\mathrm{2}×\mathrm{5}}+\frac{\alpha^{\mathrm{6}} {x}^{\mathrm{7}} }{\mathrm{3}×\mathrm{7}}−\frac{\alpha^{\mathrm{8}} {x}^{\mathrm{9}} }{\mathrm{4}×\mathrm{9}}.....\overset{\mathrm{1}} {\mid}_{\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\alpha^{\mathrm{2}} }{\mathrm{3}}−\frac{\alpha^{\mathrm{4}} }{\mathrm{10}}+\frac{\alpha^{\mathrm{6}} }{\mathrm{21}}−\frac{\alpha^{\mathrm{8}} }{\mathrm{36}}...\right) \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} {itan}^{−\mathrm{1}} \left(\alpha{x}\right)\:{dx} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/May/18

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

I_2 =∫_0 ^1 itan^(−1) αx dx =i∫_0 ^1 (αx−((α^3 x^3 )/3)+((α^5 x^5 )/5)....)dx =i∣(((αx^2 )/2)−((α^3 x^4 )/(3×4))+((α^5 x^6 )/(5×6))....)∣_0 ^1 =i((α/2)−(α^3 /(12))+(α^5 /(30))...)

$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} {itan}^{−\mathrm{1}} \alpha{x}\:{dx} \\ $$$$={i}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\alpha{x}−\frac{\alpha^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\alpha^{\mathrm{5}} {x}^{\mathrm{5}} }{\mathrm{5}}....\right){dx} \\ $$$$={i}\mid\left(\frac{\alpha{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\alpha^{\mathrm{3}} {x}^{\mathrm{4}} }{\mathrm{3}×\mathrm{4}}+\frac{\alpha^{\mathrm{5}} {x}^{\mathrm{6}} }{\mathrm{5}×\mathrm{6}}....\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={i}\left(\frac{\alpha}{\mathrm{2}}−\frac{\alpha^{\mathrm{3}} }{\mathrm{12}}+\frac{\alpha^{\mathrm{5}} }{\mathrm{30}}...\right) \\ $$

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