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Question Number 190700 by Rupesh123 last updated on 09/Apr/23

Answered by 07049753053 last updated on 09/Apr/23

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)ln\left(x\right)}{x}dx$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathbf{a}=1}{\int }_0^{\mathrm{\infty }}\frac{\mathbf{sin}\left(\mathbf{x}\right){\mathbf{x}}^{\mathbf{a}}}{\mathbf{x}}\mathbf{dx}=\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathbf{a}=1}{\int }_0^{\mathrm{\infty }}\mathbf{sin}\left(\mathbf{x}\right){\mathbf{x}}^{\mathbf{a}-1}\mathbf{dx}$$$$\mathbf{from}\mathbf{euler}\mathbf{formula}$$$${\mathbf{e}}^{-\mathbf{ix}}=\mathbf{cos}\left(\mathbf{x}\right)+\mathbf{sin}\left(\mathbf{x}\right)$$$$\mathbf{sin}\left(\mathbf{x}\right)=\mathcal{I}\mathit{m}\left({\mathit{e}}^{-\mathbf{ix}}\right)$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathbf{a}=1}\mathcal{I}\mathit{m}{\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathbf{ix}}{\mathbf{x}}^{\mathbf{a}-1}\mathbf{dx}$$$$\mathbf{let}\mathbf{ix}=\mathbf{u}\mathbf{x}=\frac{\mathbf{u}}{\mathbf{i}}\mathbf{dx}=\frac{\mathbf{du}}{\mathbf{i}}$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathbf{a}=1}\mathcal{I}\mathit{m}{\int }_0^{\mathrm{\infty }}{\mathbf{e}}^{-\mathbf{u}}{\left(\frac{\mathbf{u}}{\mathbf{i}}\right)}^{\mathbf{a}-1}\frac{\mathbf{du}}{\mathbf{i}}$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathit{a}=1}\mathcal{I}\mathit{m}\left(\frac{1}{{\mathbf{i}}^a}\right){\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathit{u}}{\mathbf{u}}^{\mathbf{a}-1}\mathbf{du}$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathit{a}=1}\mathcal{I}\mathit{m}\left(\frac{1}{{\mathit{i}}^a}\right){\int }_0^{\mathrm{\infty }}{\mathbf{e}}^{-\mathbf{u}}{\mathbf{u}}^{(\mathbf{a}-1+1)-1}\mathbf{du}$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathit{a}=1}\mathcal{I}\mathit{m}\left(\frac{\mathbf{\Gamma }\left(a\right)}{i^a}\right)=-\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathit{a}=1}(-\mathbf{\Gamma }\left(\mathit{a}\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right))=-\frac{1}{2}\mathbf{\Gamma }\left(\mathit{a}\right)(\mathit{\pi }\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right)+2\mathbf{sin}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right)\mathit{\psi }\left(\mathit{a}\right)){\mid }_{\mathit{a}=1}$$$$-\frac{1}{2}\mathbf{\Gamma }\left(1\right)(\mathit{\pi }\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\pi }}{2}\right)+2\mathit{\pi }\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{2}\right)\mathit{\psi }\left(1\right))=-\frac{\mathit{\gamma }\mathit{\pi }}{2}$$

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