Integration Questions

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

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

Answered by Frix last updated on 05/May/24

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

Commented by Ghisom last updated on 05/May/24

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

Commented by Frix last updated on 05/May/24

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