Question Number 190700 by Rupesh123 last updated on 09/Apr/23
Answered by 07049753053 last updated on 09/Apr/23
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}} }{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{euler}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ix}}} =\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathcal{I}{m}}\left(\boldsymbol{{e}}^{−\boldsymbol{\mathrm{ix}}} \right) \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \boldsymbol{\mathcal{I}{m}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{\mathrm{ix}}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{ix}}=\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{i}}}\:\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{i}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}\:} \boldsymbol{\mathcal{I}{m}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{u}}} \left(\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{i}}}\right)^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{i}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{i}}^{{a}} }\right)\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{u}}} \boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{a}}−\mathrm{1}} \boldsymbol{\mathrm{du}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\mathrm{1}}{\boldsymbol{{i}}^{{a}} }\right)\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{u}}} \boldsymbol{\mathrm{u}}^{\left(\boldsymbol{\mathrm{a}}−\mathrm{1}+\mathrm{1}\right)−\mathrm{1}} \boldsymbol{\mathrm{du}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \:\boldsymbol{\mathcal{I}{m}}\left(\frac{\boldsymbol{\Gamma}\left({a}\right)}{{i}^{{a}} }\right)=−\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left(−\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\right)=−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\left(\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)+\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\boldsymbol{\psi}\left(\boldsymbol{{a}}\right)\right)\mid_{\boldsymbol{{a}}=\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\Gamma}\left(\mathrm{1}\right)\left(\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)+\mathrm{2}\boldsymbol{\pi{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\boldsymbol{\psi}\left(\mathrm{1}\right)\right)=−\frac{\boldsymbol{\gamma\pi}}{\mathrm{2}} \\ $$