Question Number 61388 by maxmathsup by imad last updated on 02/Jun/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left({lnx}\right)}{{lnx}}\:{dx}\:. \\ $$
Commented by perlman last updated on 02/Jun/19
![let u=ln(x) ==> dx=e^u du =∫_(−∞) ^0 e^u sin(u)(du/u) let f(t)=∫_(−∞) ^0 ((e^(ut) sin(u))/u)du t>0 f′(t)=∫_(−∞) ^0 e^(ut) sin(u)dt f′(t)=[e^(ut) (−cos(t))]+t∫e^(ut) cos(t)dt =−1+t[ [e^(ut) sin(t)]−t∫e^(ut) sin(t)dt] =−1 −t^2 ∫_(−∞) ^0 e^(ut) sin(u)du=−1−t^2 f′(t) ==>f′(t)=((−1)/(1+t^2 )) f(t)=−arctg(t)+c f(0)=∫_(−∞) ^0 ((sin(t))/t)dt=(π/2) f(t)=−arctan(t)+(π/2) our integral is f(1)=−(π/4)+(π/2)=(π/4)](https://www.tinkutara.com/question/Q61407.png)
$${let}\:{u}={ln}\left({x}\right) \\ $$$$==>\:{dx}={e}^{{u}} {du} \\ $$$$=\int_{−\infty} ^{\mathrm{0}} {e}^{{u}} {sin}\left({u}\right)\frac{{du}}{{u}} \\ $$$${let}\:{f}\left({t}\right)=\int_{−\infty} ^{\mathrm{0}} \frac{{e}^{{ut}} {sin}\left({u}\right)}{{u}}{du}\:\:{t}>\mathrm{0} \\ $$$${f}'\left({t}\right)=\int_{−\infty} ^{\mathrm{0}} {e}^{{ut}} {sin}\left({u}\right){dt} \\ $$$${f}'\left({t}\right)=\left[{e}^{{ut}} \left(−{cos}\left({t}\right)\right)\right]+{t}\int{e}^{{ut}} {cos}\left({t}\right){dt} \\ $$$$=−\mathrm{1}+{t}\left[\:\:\:\left[{e}^{{ut}} {sin}\left({t}\right)\right]−{t}\int{e}^{{ut}} {sin}\left({t}\right){dt}\right] \\ $$$$=−\mathrm{1}\:−{t}^{\mathrm{2}} \int_{−\infty} ^{\mathrm{0}} {e}^{{ut}} {sin}\left({u}\right){du}=−\mathrm{1}−{t}^{\mathrm{2}} {f}'\left({t}\right) \\ $$$$==>{f}'\left({t}\right)=\frac{−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}\left({t}\right)=−{arctg}\left({t}\right)+{c} \\ $$$${f}\left(\mathrm{0}\right)=\int_{−\infty} ^{\mathrm{0}} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({t}\right)=−{arctan}\left({t}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${our}\:{integral}\:{is}\:{f}\left(\mathrm{1}\right)=−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 02/Jun/19
$${thank}\:{you}\:{sir}\: \\ $$
Commented by maxmathsup by imad last updated on 02/Jun/19
![letA=∫_0 ^1 ((sin(lnx))/(lnx))dx let use the changement ln(x)=−t ⇒x =e^(−t) A =−∫_0 ^∞ ((−sin(t))/(−t))(−e^(−t) )dt = ∫_0 ^∞ ((e^(−t) sint)/t) dt let use parametrage with f(x) =∫_0 ^∞ ((e^(−xt) sint)/t) dt with x≥0 ⇒f^′ (x) = −∫_0 ^∞ e^(−xt) sint dt =−Im(∫_0 ^∞ e^(−xt+it) dt) but ∫_0 ^∞ e^((−x+i)t) dt =[(1/(−x+i)) e^((−x+i)t) ]_(t=0) ^(+∞) =−(1/(−x +i)) =(1/(x−i)) =((x+i)/(x^2 +1)) ⇒f^′ (x) =−(1/(1+x^2 )) ⇒f(x) =−arctan(x)+λ λ =f(0) =∫_0 ^∞ ((sin(t))/t) dt =(π/2) ⇒f(x) =(π/2) −arctan(x) A =f(1) =(π/2) −arctan(1) =(π/2) −(π/4) =(π/4) .](https://www.tinkutara.com/question/Q61429.png)
$${letA}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:{let}\:{use}\:{the}\:{changement}\:{ln}\left({x}\right)=−{t}\:\Rightarrow{x}\:={e}^{−{t}} \:\: \\ $$$${A}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−{sin}\left({t}\right)}{−{t}}\left(−{e}^{−{t}} \right){dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{t}} \:\:{sint}}{{t}}\:{dt}\:\:\:{let}\:{use}\:{parametrage}\:{with} \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{xt}} \:{sint}}{{t}}\:{dt}\:\:{with}\:\:{x}\geqslant\mathrm{0}\:\Rightarrow{f}^{'} \left({x}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}} {sint}\:{dt} \\ $$$$=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}+{it}} {dt}\right)\:{but}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−{x}+{i}\right){t}} {dt}\:=\left[\frac{\mathrm{1}}{−{x}+{i}}\:{e}^{\left(−{x}+{i}\right){t}} \right]_{{t}=\mathrm{0}} ^{+\infty} \\ $$$$=−\frac{\mathrm{1}}{−{x}\:+{i}}\:=\frac{\mathrm{1}}{{x}−{i}}\:=\frac{{x}+{i}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{f}^{'} \left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)\:=−{arctan}\left({x}\right)+\lambda \\ $$$$\lambda\:={f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}\right)}{{t}}\:{dt}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({x}\right) \\ $$$${A}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by maxmathsup by imad last updated on 02/Jun/19
![anther way we have sin(u) =((e^(iu) −e^(−iu) )/(2i)) ⇒sin(ln(x))= ((e^(iln(x)) −e^(−iln(x)) )/(2i)) (a^x =e^(xln(a)) and (d/dx)(a^x )=ln(a)a^x ) ⇒sin(ln(x))=((x^i −x^(−i) )/(2iln(x))) ⇒ ∫_0 ^1 ((sin(lnx))/(lnx))dx = (1/(2i)) ∫_0 ^1 ((x^i −x^(−i) )/(ln(x)))dx let ϕ(t) =(1/(2i)) ∫_0 ^1 ((x^(it) −x^(−i) )/(ln(x)))dx we have ϕ^′ (t) =(1/(2i)) ∫_0 ^1 (∂/∂t){ ((x^(it) −x^(−i) )/(ln(x)))}dx (x^(it) =e^(itlnx)) ⇒(x^(it) )_t ′ =iln(x)x^(it) ) ⇒ϕ^′ (t) =(1/(2i)) ∫_0 ^1 ((iln(x)x^(it) )/(ln(x)))dx =(1/2) ∫_0 ^1 x^(it) dx =(1/2)[(1/(it +1)) x^(it+1) ]_(x=0) ^1 =(1/2){ (1/(it +1))} ⇒ϕ(t) =(1/2) ∫ (dt/(it +1)) +k =(1/(2i))ln(1+it) +k we have ϕ(−1) =0 =(1/(2i))ln(1−it)+k ⇒k=−(1/(2i))ln(1−it) ⇒ ϕ(t)=(1/(2i))ln(1+it)−(1/(2i))ln(1−it) ∫_0 ^1 ((sin(lnx))/(ln(x))) dx =ϕ(1) =(1/(2i))ln(((1+i)/(1−i))) 1+i =(√2)e^((iπ)/4) and 1−i =(√2)e^(−((iπ)/4)) ⇒((1+i)/(1−i)) =e^(i(π/2)) ⇒(1/(2i))ln(((1+i)/(1−i)))=(1/(2i))(((iπ)/2)) =(π/4) . ⇒ ∫_0 ^1 ((sin(lnx))/(ln(x)))dx =(π/4) .](https://www.tinkutara.com/question/Q61432.png)
$${anther}\:{way}\:\:{we}\:{have}\:{sin}\left({u}\right)\:=\frac{{e}^{{iu}} −{e}^{−{iu}} }{\mathrm{2}{i}}\:\Rightarrow{sin}\left({ln}\left({x}\right)\right)=\:\frac{{e}^{{iln}\left({x}\right)} −{e}^{−{iln}\left({x}\right)} }{\mathrm{2}{i}} \\ $$$$\left({a}^{{x}} \:={e}^{{xln}\left({a}\right)} \:\:\:{and}\:\:\frac{{d}}{{dx}}\left({a}^{{x}} \right)={ln}\left({a}\right){a}^{{x}} \right)\:\Rightarrow{sin}\left({ln}\left({x}\right)\right)=\frac{{x}^{{i}} −{x}^{−{i}} }{\mathrm{2}{iln}\left({x}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{i}} −{x}^{−{i}} }{{ln}\left({x}\right)}{dx}\:\:{let}\:\:\varphi\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{it}} −{x}^{−{i}} }{{ln}\left({x}\right)}{dx}\: \\ $$$${we}\:{have}\:\varphi^{'} \left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\partial}{\partial{t}}\left\{\:\frac{{x}^{{it}} −{x}^{−{i}} }{{ln}\left({x}\right)}\right\}{dx} \\ $$$$\left({x}^{{it}} \:={e}^{\left.{itlnx}\right)} \:\Rightarrow\left({x}^{{it}} \right)_{{t}} '\:={iln}\left({x}\right){x}^{{it}} \right)\:\Rightarrow\varphi^{'} \left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{iln}\left({x}\right){x}^{{it}} }{{ln}\left({x}\right)}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{it}} \:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{it}\:+\mathrm{1}}\:{x}^{{it}+\mathrm{1}} \right]_{{x}=\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{{it}\:+\mathrm{1}}\right\}\:\Rightarrow\varphi\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{{dt}}{{it}\:+\mathrm{1}}\:+{k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}+{it}\right)\:+{k}\:\:\:{we}\:{have}\:\:\varphi\left(−\mathrm{1}\right)\:=\mathrm{0}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}−{it}\right)+{k}\:\Rightarrow{k}=−\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}−{it}\right)\:\Rightarrow \\ $$$$\varphi\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}+{it}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}−{it}\right)\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({lnx}\right)}{{ln}\left({x}\right)}\:{dx}\:=\varphi\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\right) \\ $$$$\mathrm{1}+{i}\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\:{and}\:\:\:\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\:={e}^{{i}\frac{\pi}{\mathrm{2}}} \:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{{i}\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{4}}\:.\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({lnx}\right)}{{ln}\left({x}\right)}{dx}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by perlman last updated on 02/Jun/19
$${y}'{re}\:{welcom} \\ $$