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Question Number 205163 by Lindemann last updated on 11/Mar/24

$${\int }_0^1\frac{sin\left(lnx\right)}{lnx}dx$$

Answered by mathzup last updated on 11/Mar/24

$$I={\int }_0^1\frac{sin\left(lnx\right)}{lnx}dxchangementx=e^{-t}$$$$giveI=-{\int }_0^{\mathrm{\infty }}\frac{-sint}{-t}(-e^{-t})dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t}sintdt$$$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-at}}{t}sintdt(a>0)$$$$f^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-at}sintdt$$$$=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{-at+it}dt\right)$$$$but{\int }_0^{\mathrm{\infty }}{e}^{(-a+i)t}dt={\left[\frac{1}{-a+i}{e}^{(-a+i)t}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{1}{-a+i}=\frac{1}{a-i}=\frac{a+i}{a^2+1}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=-\frac{1}{a^2+1}\Rightarrow f\left(a\right)=c-arctana$$$$lafonctionestprolongeableparcontinuiteen0$$$$\Rightarrow \mathrm{\exists }m>0\mid e^{-at}\frac{sint}{t}\mid ⩽{me}^{-at}\Rightarrow$$$$\mid f\left(a\right)\mid ⩽m{\int }_0^{\mathrm{\infty }}{e}^{-at}dt=\frac{m}{a}\to 0(a\to +\mathrm{\infty })$$$$0=c-\frac{\pi }{2}\Rightarrow c=\frac{\pi }{2}\Rightarrow f\left(a\right)=\frac{\pi }{2}-arctan\left(a\right)$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t}sintdt=f\left(1\right)=\frac{\pi }{2}-arctan\left(1\right)$$$$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$$$so{\int }_0^1\frac{sin\left(lnx\right)}{lnx}dx=\frac{\pi }{4}$$$$$$

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