Question Number 207054 by Ghisom last updated on 05/May/24 | ||
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$$\Omega_{\alpha} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\alpha} \sqrt{−{x}\mathrm{ln}\:{x}}\:{dx}=? \\ $$ | ||
Answered by Frix last updated on 05/May/24 | ||
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$$\alpha>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\alpha} \sqrt{−{x}\mathrm{ln}\:{x}}{dx}\:\overset{{t}=−\mathrm{ln}\:{x}} {=}\:−\underset{\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{−\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right){t}} \sqrt{{t}}{dt}= \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\left(\alpha+\frac{\mathrm{3}}{\mathrm{2}}\right){t}} {dt}\:\overset{{u}=\frac{\left(\mathrm{2}\alpha+\mathrm{3}\right){t}}{\mathrm{2}}} {=} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\underset{\mathrm{0}} {\overset{\infty} {\int}}{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−{u}} {du}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\Gamma\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\Omega_{\alpha} =\frac{\sqrt{\mathrm{2}\pi}}{\left(\mathrm{2}\alpha+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$ | ||
Commented by Ghisom last updated on 05/May/24 | ||
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$$\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{substitution}\:\mathrm{but}\:\mathrm{I}\:\mathrm{did}\:\mathrm{not} \\ $$$$\mathrm{see}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{the}\:\Gamma−\mathrm{function} \\ $$ | ||
Commented by Frix last updated on 05/May/24 | ||
I remember this "trick" from long ago because it surprised me... | ||