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Question Number 131673 by liberty last updated on 07/Feb/21

$$\mathrm{Nice}\mathrm{integral}$$$${\int }_0^1\frac{\sin \left(\ln \mathrm{x}\right)}{\ln \mathrm{x}}\mathrm{dx}=?$$

Answered by mindispower last updated on 07/Feb/21

$$ln\left(x\right)=-t$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}{e}^{-t}dt=A$$$$g\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)e^{-at}}{t}dt$$$$g^{\prime }\left(a\right)=Im-{\int }_0^{\mathrm{\infty }}{e}^{(i-a)t}dt$$$$=-Im\left[\frac{1}{(a-i)}\right]=\frac{1}{a^2+1}$$$$g\left(a\right)={tan}^{-}\left(a\right)+c$$$$g\left(0\right)=\frac{\pi }{2}\Rightarrow g\left(a\right)={tan}^{-}\left(a\right)+\frac{\pi }{2}$$$$A=g\left(1\right)=\frac{\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{4}$$

Commented by liberty last updated on 07/Feb/21

$$\mathrm{i}\mathrm{think}\mathrm{is}\frac{\pi }{4}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 07/Feb/21

$$\mathrm{sir}\mathrm{small}\mathrm{error}-\mathrm{Im}\left(\frac{1}{\mathrm{a}-\mathrm{i}}\right)=-\mathrm{Im}\left(\frac{\mathrm{a}+\mathrm{i}}{1+{\mathrm{a}}^2}\right)=-\frac{1}{1+{\mathrm{a}}^2}\Rightarrow ....$$

Answered by Dwaipayan Shikari last updated on 07/Feb/21

$${\int }_0^1\frac{sin\left(log\left(x\right)\right)}{log\left(x\right)}dxlog\left(x\right)=u\Rightarrow 1=x\frac{du}{dx}$$$$={\int }_{-\mathrm{\infty }}^0\frac{sin\left(u\right)}{u}{e}^{u}duu=-j$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(i\right)}{j}{e}^{-j}duI\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{sinj}{j}{e}^{-\alpha j}dj$$$$I^{\prime }\left(\alpha \right)=-{\int }_0^{\mathrm{\infty }}sin\left(j\right)e^{-\alpha j}dj=-\frac{1}{{\alpha }^2+1}$$$$I\left(\alpha \right)=-{tan}^{-1}\alpha +C(\alpha \to \mathrm{\infty }C=\frac{\pi }{2})$$$$I\left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

Commented by liberty last updated on 07/Feb/21

$$\mathrm{okay}$$

Answered by liberty last updated on 07/Feb/21

$$\mathrm{I}\left(\beta \right)={\int }_0^1\frac{\sin \left(\beta \ln \mathrm{x}\right)}{\ln \mathrm{x}}\mathrm{dx}$$$$\mathrm{I}{}^{\prime }\left(\beta \right)={\int }_0^1\ln \left(\mathrm{x}\right)\frac{\cos \left(\beta \ln \mathrm{x}\right)}{\ln \left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{I}{}^{\prime }\left(\beta \right)={\int }_0^1\cos \left(\beta \ln \mathrm{x}\right)\mathrm{dx}$$$$\mathrm{change}\mathrm{of}\mathrm{variables}\mathrm{u}=\ln \left(\mathrm{x}\right)$$$$\mathrm{I}{}^{\prime }\left(\beta \right)=\underset{-\mathrm{\infty }}{\stackrel{0}{\int }}\cos \left(\beta \mathrm{u}\right){\mathrm{e}}^{\mathrm{u}}\mathrm{du}$$$$\mathrm{I}{}^{\prime }\left(\beta \right)=\frac{1}{1+{\beta }^2}\mathrm{then}\mathrm{I}\left(\beta \right)=\arctan \left(\beta \right)+\mathrm{C}$$$$\mathrm{setting}\beta =0\mathrm{hence}\mathrm{C}=0$$$$\mathrm{I}(\beta =1)=\arctan 1=\frac{\pi }{4}$$

Answered by mathmax by abdo last updated on 07/Feb/21

$$\mathrm{\Phi }={\int }_0^1\frac{\sin \left(\mathrm{lnx}\right)}{\mathrm{lnx}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{lnx}=-\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{e}}^{-\mathrm{t}}$$$$\mathrm{\Phi }=-{\int }_0^{\mathrm{\infty }}\frac{\sin (-\mathrm{t})}{-\mathrm{t}}(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mathrm{let}\mathrm{consider}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{at}}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\mathrm{sint}{\mathrm{e}}^{-\mathrm{at}}\mathrm{dt}$$$$=-\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{it}-\mathrm{at}}\mathrm{dt}\right)\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{a}+\mathrm{i})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{-\mathrm{a}+\mathrm{i}}{\mathrm{e}}^{(-\mathrm{a}+\mathrm{i})\mathrm{t}}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{1}{\mathrm{a}-\mathrm{i}}(-1)=\frac{1}{\mathrm{a}-\mathrm{i}}=\frac{\mathrm{a}+\mathrm{i}}{{\mathrm{a}}^2+1}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{1}{1+{\mathrm{a}}^2}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\mathrm{c}-\arctan \left(\mathrm{a}\right)$$$$\mathrm{f}\left(0\right)=\mathrm{c}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}=\frac{\pi }{2}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}-\arctan \left(\mathrm{a}\right)$$$$\mathrm{\Phi }=\mathrm{f}\left(1\right)=\frac{\pi }{2}-\arctan \left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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