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Question Number 85721 by M±th+et£s last updated on 24/Mar/20

$$showthat$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-x}ln\left(x\right)}{\sqrt{x}}dx=-\sqrt{\pi }(\gamma +ln\left(4\right))$$

Answered by mind is power last updated on 24/Mar/20

$${\int }_0^{+\mathrm{\infty }}\frac{e^{-x}ln\left(x\right)}{\sqrt{x}}dx$$$$\sqrt{x}=u$$$$\Rightarrow {\int }_0^{+\mathrm{\infty }}4e^{-u^2}ln\left(u\right)du=4{\int }_0^{+\mathrm{\infty }}{e}^{-u^2}ln\left(u\right)du$$$${\int }_0^{+\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt=\mathrm{\Gamma }\left(x\right)$$$$t=u^2\Rightarrow dt=2udu$$$$2{\int }_0^{+\mathrm{\infty }}{u}^{2x-1}{e}^{-u^2}=\mathrm{\Gamma }\left(x\right)\Rightarrow {\mathrm{\Gamma }}^{\prime }\left(x\right)=2\int 2ln\left(u\right)u^{2x-1}{e}^{-u^2}du={\mathrm{\Gamma }}^{\prime }\left(x\right)$$$$\Rightarrow 4{\int }_0^{+\mathrm{\infty }}ln\left(u\right)e^{-u^2}du={\mathrm{\Gamma }}^{\prime }\left(\frac{1}{2}\right)=\mathrm{\Psi }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$\mathrm{\Psi }\left(\frac{1}{2}\right)=-\gamma -ln(2.2)=-\gamma -ln\left(4\right)$$$$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }\Rightarrow$$$${\int }_0^{+\mathrm{\infty }}\frac{e^{-x}ln\left(x\right)}{\sqrt{x}}dx=4{\int }_0^{+\mathrm{\infty }}ln\left(u\right)e^{-u^2}du=\sqrt{\pi }(-\gamma -ln\left(4\right))=-\sqrt{\pi }(\gamma +ln\left(4\right))$$

Commented by M±th+et£s last updated on 24/Mar/20

$$godblessyousir$$

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