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Question Number 163487 by mnjuly1970 last updated on 07/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{{sin}^2\left(\ln \left(x\right)\right).\ln \left(x\right)}{\sqrt{x}}dx=?$$$$-----$$

Answered by Lordose last updated on 07/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\frac{1}{2}-1}{\sin }^2\left(\ln \left(\mathrm{x}\right)\right)\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{x}={\mathrm{e}}^{\mathrm{x}}}{=}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{xe}}^{\frac{\mathrm{x}}{2}}{\sin }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{\Omega }=\mathfrak{Im}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{xe}}^{\frac{\mathrm{x}}{2}}\cdot {\mathrm{e}}^{-2\mathrm{ix}}\mathrm{dx}$$$$\mathrm{\Omega }=\mathfrak{Im}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{xe}}^{-\mathrm{x}(2\mathbf{i}-\frac{1}{2})}\mathrm{dx}\stackrel{\mathrm{x}=\frac{\mathrm{y}}{2\mathbf{i}-\frac{1}{2}}}{=}\mathfrak{Im}\left(\frac{1}{{(2\mathbf{i}-\frac{1}{2})}^2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{ye}}^{-\mathrm{y}}\mathrm{dy}\right)$$$$\mathrm{\Omega }=\mathrm{Diverges}$$

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