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Question Number 150747 by mnjuly1970 last updated on 15/Aug/21

$$$$$$prove::$$$$\mathrm{\Omega }:={\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{x}}.ln\left(\sqrt[4]{x}\right)\stackrel{?}{=}1-\gamma$$$$m.n..$$

Answered by Olaf_Thorendsen last updated on 15/Aug/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{x}}\ln \left(\sqrt[4]{x}\right)dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-u}\ln \left(\sqrt{u}\right)\left(2udu\right)$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{ue}^{-u}\ln udu$$$$\psi \left(z\right)=\frac{{\int }_0^{\mathrm{\infty }}{y}^{z-1}\ln y{e}^{-y}dy}{{\int }_0^{\mathrm{\infty }}{y}^{z-1}{e}^{-y}dy}$$$$\psi \left(2\right)=\frac{{\int }_0^{\mathrm{\infty }}y.\ln y{e}^{-y}dy}{{\int }_0^{\mathrm{\infty }}{ye}^{-y}dy}=\frac{\mathrm{\Omega }}{{\int }_0^{\mathrm{\infty }}{ye}^{-y}dy}$$$${\int }_0^{\mathrm{\infty }}{ue}^{-u}du={[-{ue}^{-u}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{e}^{-u}du=1$$$$\Rightarrow \mathrm{\Omega }=\psi \left(2\right)$$$$$$$$\psi (z+1)=\psi \left(z\right)+\frac{1}{z}$$$$\psi \left(2\right)=\psi \left(1\right)+\frac{1}{1}=-\gamma +1$$$$$$$$\mathrm{\Omega }=1-\gamma =\psi \left(2\right)$$

Commented by mnjuly1970 last updated on 16/Aug/21

$$gratefulmrolaf..$$

Answered by mathmax by abdo last updated on 17/Aug/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}\ln \left({\mathrm{x}}^{\frac{1}{4}}\right)\mathrm{dx}\Rightarrow \mathrm{\Psi }{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\ln \left({\mathrm{t}}^{\frac{1}{2}}\right)\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{2}\ln \left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}{\mathrm{te}}^{-\mathrm{t}}\ln \left(\mathrm{t}\right)\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\left(\mathrm{tlnt}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mathrm{by}\mathrm{parts}\mathrm{u}=\mathrm{tlnt}\mathrm{and}{\mathrm{v}}^{{}^{\prime }}={\mathrm{e}}^{-\mathrm{t}}$$$$={[-{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{tlnt}\right)]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}(\mathrm{lnt}+1)(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}$$$$=0+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}(1+\mathrm{lnt})\mathrm{dt}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{lnt}\mathrm{dt}$$$$=1-\gamma$$

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