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Question Number 162924 by mnjuly1970 last updated on 02/Jan/22

$$$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}.\ln \left(x\right)}{\sqrt{x}}dx=\lambda \mathrm{\Gamma }\left(\frac{1}{4}\right)$$$$$$$$\lambda =?◼$$$$$$

Answered by Lordose last updated on 02/Jan/22

$$\varnothing ={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-\frac{1}{2}}{\mathrm{e}}^{-{\mathrm{x}}^2}\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$\varnothing \stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{2}}}{=}\frac{1}{4}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-\frac{1}{4}+\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{x}}\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$\varnothing \left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}-1}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=\mathbf{\Gamma }\left(\mathrm{a}\right)$$$${\varnothing }^{\prime }\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}-1}{\mathrm{e}}^{-\mathrm{x}}\ln \left(\mathrm{x}\right)\mathrm{dx}=\mathbf{\Gamma }\left(\mathrm{a}\right){\mathit{\psi }}^{\left(0\right)}\left(\mathrm{a}\right)$$$$\varnothing ={\varnothing }^{\prime }\left(\frac{1}{4}\right)=\mathbf{\Gamma }\left(\frac{1}{4}\right){\mathit{\psi }}^{\left(0\right)}\left(\frac{1}{4}\right)$$$$\lambda ={\mathit{\psi }}^{\left(0\right)}\left(\frac{1}{4}\right)$$

Commented by mnjuly1970 last updated on 02/Jan/22

$$thanksalotsirlordose$$

Answered by mathmax by abdo last updated on 02/Jan/22

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{lnx}}{\sqrt{\mathrm{x}}}\mathrm{dx}\Rightarrow \mathrm{\Psi }{=}_{\mathrm{x}=\sqrt{\mathrm{t}}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}\ln \left(\sqrt{\mathrm{t}}\right)}{{\mathrm{t}}^{\frac{1}{4}}}\frac{1}{2}{\mathrm{t}}^{-\frac{1}{2}}\mathrm{dt}$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{lnt}{\mathrm{t}}^{-\frac{3}{4}}\mathrm{dt}=\frac{1}{4}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{4}-1}{\mathrm{e}}^{-\mathrm{t}}\ln \left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{let}\mathrm{\Gamma }\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{\Gamma }}^{\left(1\right)}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-1}\mathrm{lnt}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\Rightarrow$$$${\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{1}{4}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{4}-1}{\mathrm{e}}^{-\mathrm{t}}\ln \left(\mathrm{t}\right)\mathrm{dt}=\psi \left(\frac{1}{4}\right)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{lnx}}{\sqrt{\mathrm{x}}}\mathrm{dx}=\frac{1}{4}\psi \left(\frac{1}{4}\right)$$

Commented by mathmax by abdo last updated on 03/Jan/22

$$\mathrm{sorry}\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}=\psi \left(\frac{1}{4}\right)\Rightarrow {\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{1}{4}\right)=\psi \left(\frac{1}{4}\right).\mathrm{\Gamma }\left(\frac{1}{4}\right)$$

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