Question Number 150655 by mnjuly1970 last updated on 14/Aug/21
Answered by Ar Brandon last updated on 14/Aug/21
$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\sqrt{{x}}} \mathrm{ln}\left(\sqrt{{x}}\right)}{{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }{dx}\underset{{x}={u}^{\mathrm{2}} } {=}\mathrm{2}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} \mathrm{ln}{u}\:{du} \\ $$$$\:\:=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} {u}^{\alpha} {e}^{−{u}} {du}=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \Gamma\left(\alpha+\mathrm{1}\right) \\ $$$$\:\:=\mathrm{2}\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\sqrt{\pi}\left(\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:=\sqrt{\pi}\left(\mathrm{2}−\gamma−\mathrm{2ln2}\right)=\sqrt{\pi}\left(\mathrm{2}−\gamma−\mathrm{ln4}\right) \\ $$
Commented by mnjuly1970 last updated on 14/Aug/21
$$\:{thamk}\:{you}\:{so}\:{much}\:{mr}\:{brandon}… \\ $$
Commented by Ar Brandon last updated on 14/Aug/21
My pleasure, Sir