Question Number 47677 by vajpaithegrate@gmail.com last updated on 13/Nov/18
$$\int\mathrm{x}^{\mathrm{x}\:} \mathrm{dx}= \\ $$
Commented by maxmathsup by imad last updated on 13/Nov/18
$${at}\:{form}\:{of}\:{serie}\:\: \\ $$$$\int\:{x}^{{x}} {dx}\:=\int\:{e}^{{xln}\left({x}\right)} {dx}=\int\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} \left({lnx}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:{x}^{{n}} \left({lnx}\right)^{{n}} {dx}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{A}_{{n}} }{{n}!}\:\:{with}\:{A}_{{n}} =\:\int\:{x}^{{n}} \left({ln}\left({x}\right)\right)^{{n}} {dx} \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \left({lnx}\right)^{{n}} \:−\int\:\frac{{x}^{{n}} }{{n}+\mathrm{1}}{n}\left({lnx}\right)^{{n}−\mathrm{1}} {dx} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \left({ln}\left({x}\right)\right)^{{n}} \:−\frac{{n}}{{n}+\mathrm{1}}\:\int\:\:{x}^{{n}} \left({lnx}\right)^{{n}−\mathrm{1}} {dx}\:\:{and}\:{we}\:{can}\:{determine}\:{A}_{{n}} \\ $$$${if}\:{we}\:{have}\:{the}\:{limits}. \\ $$
Commented by vajpaithegrate@gmail.com last updated on 13/Nov/18
$$\mathrm{tnq}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 13/Nov/18
$${you}\:{are}\:{welcome}. \\ $$