Question Number 64224 by mmkkmm000m last updated on 16/Jul/19 | ||
$$\int{x}_{{x}} {dx} \\ $$$$ \\ $$$$\int{x}^{{x}} {dx} \\ $$ | ||
Commented by mathmax by abdo last updated on 17/Jul/19 | ||
$${I}\:=\int\:{x}^{{x}} {dx}\:=\int\:\:{e}^{{xlnx}} {dx}\:=\int\:\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} \left({lnx}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:\:{x}^{{n}} \left({lnx}\right)^{{n}} \:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{w}_{{n}} }{{n}!}\:\:{with}\:{w}_{{n}} =\int\:{x}^{{n}} \left({lnx}\right)^{{n}} {dx} \\ $$$${changement}\:{lnx}\:={t}\:{give}\:{x}={e}^{{t}} \:\Rightarrow{w}_{{n}} =\int\:{e}^{{nt}} \:{t}^{{n}} {e}^{{t}} {dt} \\ $$$$=\int\:\:{t}^{{n}} \:{e}^{\left({n}+\mathrm{1}\right){t}} {dt}\:=_{\left({n}+\mathrm{1}\right){t}\:={u}} \:\:\:\:\int\:\:\left(\frac{{u}}{{n}+\mathrm{1}}\right)^{{n}} \:{e}^{{u}} \:\frac{{du}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int\:\:{u}^{{n}} \:{e}^{{u}} \:{du}\:\Rightarrow\:{I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int\:\:{u}^{{n}} \:{e}^{{n}} {du}\:\:+{c} \\ $$$$ \\ $$ | ||