Question Number 85721 by M±th+et£s last updated on 24/Mar/20
$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\:\sqrt{{x}}}{dx}=−\sqrt{\pi}\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$
Answered by mind is power last updated on 24/Mar/20
$$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\:\sqrt{{x}}}{dx} \\ $$$$\sqrt{{x}}={u} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \mathrm{4}{e}^{−{u}^{\mathrm{2}} } {ln}\left({u}\right){du}=\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {ln}\left({u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{+\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}=\Gamma\left({x}\right) \\ $$$${t}={u}^{\mathrm{2}} \Rightarrow{dt}=\mathrm{2}{udu} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {u}^{\mathrm{2}{x}−\mathrm{1}} {e}^{−{u}^{\mathrm{2}} } =\Gamma\left({x}\right)\Rightarrow\Gamma'\left({x}\right)=\mathrm{2}\int\mathrm{2}{ln}\left({u}\right){u}^{\mathrm{2}{x}−\mathrm{1}} {e}^{−{u}^{\mathrm{2}} } {du}=\Gamma'\left({x}\right) \\ $$$$\Rightarrow\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {ln}\left({u}\right){e}^{−{u}^{\mathrm{2}} } {du}=\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−{ln}\left(\mathrm{2}.\mathrm{2}\right)=−\gamma−{ln}\left(\mathrm{4}\right) \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} {ln}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\mathrm{4}\int_{\mathrm{0}} ^{+\infty} {ln}\left({u}\right){e}^{−{u}^{\mathrm{2}} } {du}=\sqrt{\pi}\left(−\gamma−{ln}\left(\mathrm{4}\right)\right)=−\sqrt{\pi}\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$
Commented by M±th+et£s last updated on 24/Mar/20
$${god}\:{bless}\:{you}\:{sir} \\ $$