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Question Number 207054 by Ghisom last updated on 05/May/24

Ω_α =∫_0 ^1 x^α (√(−xln x)) dx=?

$$\Omega_{\alpha} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\alpha} \sqrt{−{x}\mathrm{ln}\:{x}}\:{dx}=? \\ $$

Answered by Frix last updated on 05/May/24

α>−(3/2) ∫_0 ^1 x^α (√(−xln x))dx =^(t=−ln x) −∫_∞ ^0 e^(−(α+(3/2))t) (√t)dt= =∫_0 ^∞ t^(1/2) e^(−(α+(3/2))t) dt =^(u=(((2α+3)t)/2)) =((2(√2))/((2α+3)^(3/2) ))∫_0 ^∞ u^(1/2) e^(−u) du=((2(√2))/((2α+3)^(3/2) ))Γ ((3/2)) ⇒ Ω_α =((√(2π))/((2α+3)^(3/2) ))

$$\alpha>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\alpha} \sqrt{−{x}\mathrm{ln}\:{x}}{dx}\:\overset{{t}=−\mathrm{ln}\:{x}} {=}\:−\underset{\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{−\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right){t}} \sqrt{{t}}{dt}= \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right){t}} {dt}\:\overset{{u}=\frac{\left(\mathrm{2}\alpha+\mathrm{3}\right){t}}{\mathrm{2}}} {=} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\underset{\mathrm{0}} {\overset{\infty} {\int}}{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−{u}} {du}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\Gamma\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\Omega_{\alpha} =\frac{\sqrt{\mathrm{2}\pi}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Commented by Ghisom last updated on 05/May/24

thank you I found the 1^(st) substitution but I did not see the shape of the Γ−function

$$\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{substitution}\:\mathrm{but}\:\mathrm{I}\:\mathrm{did}\:\mathrm{not} \\ $$$$\mathrm{see}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{the}\:\Gamma−\mathrm{function} \\ $$

Commented by Frix last updated on 05/May/24

�� I remember this "trick" from long ago because it surprised me...

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