Question Number 85256 by john santu last updated on 20/Mar/20
$$\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\frac{\mathrm{sin}\:\left(\mathrm{ln}\:\mathrm{x}\right)}{\mathrm{ln}\:\left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$
Commented by john santu last updated on 20/Mar/20
![sin (z) = ((e^(iz ) −e^(−iz) )/(2i)) sin(ln x) =((e^(i ln(x)) −e^(−i ln(x)) )/(2i)) = ((x^i −x^(−i) )/(2i)) ∫ _0 ^( 1) ((x^i −x^(−i) )/(2i ln(x))) dx = I(1) let I(b) = ∫ _0 ^( 1) ((x^(bi) −x^(−i) )/(2i ln(x))) dx I′(b) = ∫ _0 ^( 1) (∂/∂b) [((x^(bi) −x^(−i) )/(2i ln(x)))] dx I′(b) = ∫ _0 ^1 ((x^(bi) ln(x) i)/(2i ln(x))) dx I′(b) = (1/(2(1+bi))) x^(bi+1) ] _(x = 0)^(x=1) I′(b) = (1/(2(1+bi))) ⇒I(b) = ∫ (1/(2(1+bi))) db I(b) = (1/(2i)) ln(1+bi) + C](https://www.tinkutara.com/question/Q85258.png)
$$\mathrm{sin}\:\left(\mathrm{z}\right)\:=\:\frac{\mathrm{e}^{\mathrm{iz}\:} −\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2i}} \\ $$$$\mathrm{sin}\left(\mathrm{ln}\:\mathrm{x}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\:\mathrm{ln}\left(\mathrm{x}\right)} −\mathrm{e}^{−\mathrm{i}\:\mathrm{ln}\left(\mathrm{x}\right)} }{\mathrm{2i}}\:=\:\frac{\mathrm{x}^{\mathrm{i}} \:−\mathrm{x}^{−\mathrm{i}} }{\mathrm{2i}} \\ $$$$\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\frac{\mathrm{x}^{\mathrm{i}} −\mathrm{x}^{−\mathrm{i}} }{\mathrm{2i}\:\mathrm{ln}\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\mathrm{I}\left(\mathrm{1}\right) \\ $$$$\mathrm{let}\:\mathrm{I}\left(\mathrm{b}\right)\:=\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\frac{\mathrm{x}^{\mathrm{bi}} −\mathrm{x}^{−\mathrm{i}} }{\mathrm{2i}\:\mathrm{ln}\left(\mathrm{x}\right)}\:\mathrm{dx} \\ $$$$\mathrm{I}'\left(\mathrm{b}\right)\:=\:\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\frac{\partial}{\partial\mathrm{b}}\:\left[\frac{\mathrm{x}^{\mathrm{bi}} −\mathrm{x}^{−\mathrm{i}} }{\mathrm{2i}\:\mathrm{ln}\left(\mathrm{x}\right)}\right]\:\mathrm{dx} \\ $$$$\mathrm{I}'\left(\mathrm{b}\right)\:=\:\int\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\:}}\frac{\mathrm{x}^{\mathrm{bi}} \:\mathrm{ln}\left(\mathrm{x}\right)\:\mathrm{i}}{\mathrm{2i}\:\mathrm{ln}\left(\mathrm{x}\right)}\:\mathrm{dx} \\ $$$$\left.\mathrm{I}'\left(\mathrm{b}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{bi}\right)}\:\mathrm{x}^{\mathrm{bi}+\mathrm{1}} \:\right]\:_{\mathrm{x}\:=\:\mathrm{0}} ^{\mathrm{x}=\mathrm{1}} \\ $$$$\mathrm{I}'\left(\mathrm{b}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{bi}\right)}\:\Rightarrow\mathrm{I}\left(\mathrm{b}\right)\:=\:\int\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{bi}\right)}\:\mathrm{db} \\ $$$$\mathrm{I}\left(\mathrm{b}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2i}}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{bi}\right)\:+\:\mathrm{C} \\ $$
Commented by john santu last updated on 20/Mar/20
$$\frac{\mathrm{1}}{\mathrm{2i}}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{bi}\right)\:+\mathrm{C}\:=\:\int\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\:}}\:\frac{\mathrm{x}^{\mathrm{bi}} −\mathrm{x}^{−\mathrm{i}} }{\mathrm{2i}\:\mathrm{ln}\left(\mathrm{x}\right)}\:\mathrm{dx} \\ $$$$\mathrm{b}=−\mathrm{1}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2i}}\:\mathrm{ln}\left(\mathrm{1}−\mathrm{i}\right)+\mathrm{C}\:=\:\mathrm{0} \\ $$$$\mathrm{C}\:=\:−\frac{\mathrm{1}}{\mathrm{2i}}\:\mathrm{ln}\left(\mathrm{1}−\mathrm{i}\right) \\ $$$$\mathrm{I}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2i}}\:\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{i}}{\mathrm{1}−\mathrm{i}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2i}}\:\frac{\pi\mathrm{i}}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\therefore\:\int\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\:}}\:\frac{\mathrm{sin}\:\left(\mathrm{ln}\:\mathrm{x}\right)}{\mathrm{ln}\:\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 20/Mar/20
![I =∫_0 ^1 ((sin(lnx))/(lnx))dx changement lnx =−t give x =e^(−t) ⇒ I =− ∫_0 ^(+∞) ((sin(t))/t)(−e^(−t) )dt =∫_0 ^∞ ((sint)/t) e^(−t) dt let f(a) =∫_0 ^∞ ((sint)/t)e^(−at) dt with a≥0 we have f^′ (a) =−∫_0 ^∞ sint e^(−at) dt =−Im(∫_0 ^∞ e^(−at+it) dt) ∫_0 ^∞ e^((−a+i)t) dt =[(1/(−a+i)) e^((−a+i)t) ]_0 ^(+∞) =−(1/(a−i)){−1}=(1/(a−i)) =((a+i)/(a^2 +1)) ⇒f^′ (a) =−(1/(1+a^2 )) ⇒f(a) =k−arctan(a) f(0) =(π/2) =k ⇒f(a) =(π/2) −arctan(a) I =f(1) =(π/2)−(π/4) =(π/4)](https://www.tinkutara.com/question/Q85308.png)
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:\:{changement}\:{lnx}\:=−{t}\:{give}\:{x}\:={e}^{−{t}} \:\Rightarrow \\ $$$${I}\:=−\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{sin}\left({t}\right)}{{t}}\left(−{e}^{−{t}} \right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sint}}{{t}}\:{e}^{−{t}} \:{dt} \\ $$$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{{t}}{e}^{−{at}} \:{dt}\:\:{with}\:{a}\geqslant\mathrm{0}\:\:{we}\:{have} \\ $$$${f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:{sint}\:{e}^{−{at}} {dt}\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{at}+{it}} \:{dt}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{a}+{i}\right){t}} \:{dt}\:=\left[\frac{\mathrm{1}}{−{a}+{i}}\:{e}^{\left(−{a}+{i}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \:=−\frac{\mathrm{1}}{{a}−{i}}\left\{−\mathrm{1}\right\}=\frac{\mathrm{1}}{{a}−{i}} \\ $$$$=\frac{{a}+{i}}{{a}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{f}^{'} \left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }\:\Rightarrow{f}\left({a}\right)\:={k}−{arctan}\left({a}\right) \\ $$$${f}\left(\mathrm{0}\right)\:=\frac{\pi}{\mathrm{2}}\:={k}\:\Rightarrow{f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({a}\right) \\ $$$${I}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\ $$