Question Number 132459 by mnjuly1970 last updated on 14/Feb/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:{nice}\:\:\:{calculus}\:…. \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}^{\mathrm{2}} \right){ln}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}= \\ $$$$\:\:{solution}: \\ $$$$\boldsymbol{\phi}\overset{{x}^{\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right){ln}\left(\sqrt{{t}}\:\right)}{{t}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:\frac{{dt}}{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right){ln}\left({t}\right)}{{t}^{\frac{\mathrm{5}}{\mathrm{4}}} }{dt} \\ $$$$\:\:\:\boldsymbol{\psi}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right).{t}^{{a}} }{{t}^{\frac{\mathrm{5}}{\mathrm{4}}} }{dt}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}^{\frac{\mathrm{5}}{\mathrm{4}}−{a}} }{dt} \\ $$$$\:\boldsymbol{\psi}\left({a}\right)=\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}−\boldsymbol{{a}}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$$$\:\therefore\:\:\:\boldsymbol{\phi}=\frac{\boldsymbol{\psi}'\left(\mathrm{0}\right)}{\mathrm{4}}\:…… \\ $$$$\:\:\:\boldsymbol{\psi}'\left({a}\right)=\frac{\pi}{\mathrm{2}}\left[\frac{\Gamma'\left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\pi{a}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){cos}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\boldsymbol{\pi}{a}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){sin}^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\boldsymbol{\pi}{a}}{\mathrm{2}}\right)}\right] \\ $$$$\:\:\boldsymbol{\psi}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\left(\frac{\Gamma'\left(\frac{\mathrm{5}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{2}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{4}\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{2}\pi\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right) \\ $$$$\:\:\:\: \\ $$$$\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{4}}\left(\frac{\left(\mathrm{2}\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{8}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\pi\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right)\: \\ $$