Question Number 205163 by Lindemann last updated on 11/Mar/24
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({lnx}\right)}{{lnx}}{dx} \\ $$
Answered by mathzup last updated on 11/Mar/24
![I=∫_0 ^1 ((sin(lnx))/(lnx))dx changement x=e^(−t) give I=−∫_0 ^∞ ((−sint)/(−t))(−e^(−t) )dt =∫_0 ^∞ (e^(−t) /t)sint dt let f(a)=∫_0 ^∞ (e^(−at) /t)sint dt (a>0) f^′ (a)=−∫_0 ^∞ e^(−at) sint dt =−Im(∫_0 ^∞ e^(−at+it) dt) but ∫_0 ^∞ e^((−a+i)t) dt=[(1/(−a+i))e^((−a+i)t) ]_0 ^∞ =−(1/(−a+i))=(1/(a−i))=((a+i)/(a^2 +1)) ⇒ f^′ (a)=−(1/(a^2 +1)) ⇒f(a)=c−arctana la fonction est prolongeable par continuite en0 ⇒∃m>0 ∣e^(−at) ((sint)/t)∣≤me^(−at) ⇒ ∣f(a)∣≤m∫_0 ^∞ e^(−at) dt =(m/a) →0 (a→+∞) 0=c−(π/2) ⇒c=(π/2) ⇒f(a)=(π/2)−arctan(a) ∫_0 ^∞ (e^(−t) /t)sint dt =f(1)=(π/2)−arctan(1) =(π/2)−(π/4)=(π/4) so ∫_0 ^1 ((sin(lnx))/(lnx))dx=(π/4)](https://www.tinkutara.com/question/Q205169.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:{changement}\:{x}={e}^{−{t}} \\ $$$${give}\:{I}=−\int_{\mathrm{0}} ^{\infty} \:\frac{−{sint}}{−{t}}\left(−{e}^{−{t}} \right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{{t}}{sint}\:{dt} \\ $$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{at}} }{{t}}{sint}\:{dt}\:\:\:\:\:\:\left({a}>\mathrm{0}\right) \\ $$$${f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−{at}} {sint}\:{dt} \\ $$$$=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{at}+{it}} {dt}\right) \\ $$$${but}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{a}+{i}\right){t}} {dt}=\left[\frac{\mathrm{1}}{−{a}+{i}}{e}^{\left(−{a}+{i}\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{−{a}+{i}}=\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{f}\left({a}\right)={c}−{arctana} \\ $$$${la}\:{fonction}\:{est}\:{prolongeable}\:{par}\:{continuite}\:{en}\mathrm{0} \\ $$$$\Rightarrow\exists{m}>\mathrm{0}\:\mid{e}^{−{at}} \:\frac{{sint}}{{t}}\mid\leqslant{me}^{−{at}} \:\Rightarrow \\ $$$$\mid{f}\left({a}\right)\mid\leqslant{m}\int_{\mathrm{0}} ^{\infty} {e}^{−{at}} {dt}\:=\frac{{m}}{{a}}\:\rightarrow\mathrm{0}\:\:\left({a}\rightarrow+\infty\right) \\ $$$$\mathrm{0}={c}−\frac{\pi}{\mathrm{2}}\:\Rightarrow{c}=\frac{\pi}{\mathrm{2}}\:\Rightarrow{f}\left({a}\right)=\frac{\pi}{\mathrm{2}}−{arctan}\left({a}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{{t}}{sint}\:{dt}\:={f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$$${so}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$