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Question Number 102058 by Dwaipayan Shikari last updated on 06/Jul/20

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

Answered by prakash jain last updated on 06/Jul/20

$$x=e^u$$$$dx=e^{u}du$$$$I={\int }_{-\mathrm{\infty }}^0\frac{\sin u}{u}{e}^{u}du=-{\int }_0^{\mathrm{\infty }}\frac{e^{-u}\sin u}{u}du$$$$J\left(t\right)=-{\int }_0^{\mathrm{\infty }}\frac{e^{-tu}\sin u}{u}du$$$$J^{\prime }\left(t\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-tu}\sin udu=-\frac{1}{1+t^2}$$$$J\left(t\right)=-{\tan }^{-1}t+C$$$$I=J\left(1\right)$$$$\mathrm{Will}\mathrm{continue}.$$

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