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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

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:{changement}\:{x}={e}^{−{t}} \\$$$${give}\:{I}=−\int_{\mathrm{0}} ^{\infty} \:\frac{−{sint}}{−{t}}\left(−{e}^{−{t}} \right){dt} \\$$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{{t}}{sint}\:{dt} \\$$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{at}} }{{t}}{sint}\:{dt}\:\:\:\:\:\:\left({a}>\mathrm{0}\right) \\$$$${f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−{at}} {sint}\:{dt} \\$$$$=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{at}+{it}} {dt}\right) \\$$$${but}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{a}+{i}\right){t}} {dt}=\left[\frac{\mathrm{1}}{−{a}+{i}}{e}^{\left(−{a}+{i}\right){t}} \right]_{\mathrm{0}} ^{\infty} \\$$$$=−\frac{\mathrm{1}}{−{a}+{i}}=\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\$$$${f}^{'} \left({a}\right)=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{f}\left({a}\right)={c}−{arctana} \\$$$${la}\:{fonction}\:{est}\:{prolongeable}\:{par}\:{continuite}\:{en}\mathrm{0} \\$$$$\Rightarrow\exists{m}>\mathrm{0}\:\mid{e}^{−{at}} \:\frac{{sint}}{{t}}\mid\leqslant{me}^{−{at}} \:\Rightarrow \\$$$$\mid{f}\left({a}\right)\mid\leqslant{m}\int_{\mathrm{0}} ^{\infty} {e}^{−{at}} {dt}\:=\frac{{m}}{{a}}\:\rightarrow\mathrm{0}\:\:\left({a}\rightarrow+\infty\right) \\$$$$\mathrm{0}={c}−\frac{\pi}{\mathrm{2}}\:\Rightarrow{c}=\frac{\pi}{\mathrm{2}}\:\Rightarrow{f}\left({a}\right)=\frac{\pi}{\mathrm{2}}−{arctan}\left({a}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{{t}}{sint}\:{dt}\:={f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{1}\right) \\$$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\$$$${so}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}=\frac{\pi}{\mathrm{4}} \\$$$$\\$$