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Question Number 150655 by mnjuly1970 last updated on 14/Aug/21

Answered by Ar Brandon last updated on 14/Aug/21

I=∫_0 ^∞ ((e^(−(√x)) ln((√x)))/x^(1/4) )dx=_(x=u^2 ) 2∫_0 ^∞ u^(1/2) e^(−u) lnu du    =2(∂/∂α)∣_(α=(1/2)) ∫_0 ^∞ u^α e^(−u) du=2(∂/∂α)∣_(α=(1/2)) Γ(α+1)    =2Γ′((3/2))=2Γ((3/2))ψ((3/2))=(√π)(2+ψ((1/2)))    =(√π)(2−γ−2ln2)=(√π)(2−γ−ln4)

$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\sqrt{{x}}} \mathrm{ln}\left(\sqrt{{x}}\right)}{{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }{dx}\underset{{x}={u}^{\mathrm{2}} } {=}\mathrm{2}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} \mathrm{ln}{u}\:{du} \\ $$$$\:\:=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} {u}^{\alpha} {e}^{−{u}} {du}=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \Gamma\left(\alpha+\mathrm{1}\right) \\ $$$$\:\:=\mathrm{2}\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\sqrt{\pi}\left(\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:=\sqrt{\pi}\left(\mathrm{2}−\gamma−\mathrm{2ln2}\right)=\sqrt{\pi}\left(\mathrm{2}−\gamma−\mathrm{ln4}\right) \\ $$

Commented by mnjuly1970 last updated on 14/Aug/21

 thamk you so much mr brandon...

$$\:{thamk}\:{you}\:{so}\:{much}\:{mr}\:{brandon}... \\ $$

Commented by Ar Brandon last updated on 14/Aug/21

My pleasure, Sir

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