Question Number 207906 by nachosam last updated on 30/May/24
$${help} \\ $$$$\int_{\mathrm{1}} ^{\:\infty} {x}^{−{ln}\left({x}\right)} {dx} \\ $$$$ \\ $$
Answered by Berbere last updated on 30/May/24
$$\int_{\mathrm{0}} ^{\infty} \left({e}^{{t}} \right)^{\left(−{t}\right)} {e}^{{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} +{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}} {dt} \\ $$$$={e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}={e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}+{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{0}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${A}={e}^{\frac{\mathrm{1}}{\mathrm{4}}} .\frac{\sqrt{\pi}}{\mathrm{2}}−{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${erf}\left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}\Rightarrow{erf}\left(−{x}\right)=−{erf}\left({x}\right) \\ $$$${A}=\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{erf}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$