Question Number 62906 by Prithwish sen last updated on 26/Jun/19
$$\int\mathrm{x}^{\mathrm{x}} \mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
$${let}\:{A}\:=\int\:{x}^{{x}} \:{dx}\:\Rightarrow{A}\:=\int\:{e}^{{xln}\left({x}\right)} {dx}\:=\int\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} \left({ln}\left({x}\right)\right)^{{n}} }{{n}!}\:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:{x}^{{n}} \left({ln}\left({x}\right)\right)^{{n}} \:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{W}_{{n},{n}} }{{n}!}\:\:\:{with}\:{W}_{{n},{p}} =\int\:{x}^{{n}} \left({lnx}\right)^{{p}} \:{dx} \\ $$$${by}\:{psrts}\:{W}_{{n},{p}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \left({ln}\left({x}\right)\right)^{{p}} \:−\int\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:\frac{{p}}{{x}}\:\left({ln}\left({x}\right)\right)^{{p}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} \left({ln}\left({x}\right)\right)^{{p}} \:−\frac{{p}}{{n}+\mathrm{1}}\:\int\:{x}^{{n}} \:\left({lnx}\right)^{{p}−\mathrm{1}} \:{dx}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \left({lnx}\right)^{{p}} −\frac{{p}}{{n}+\mathrm{1}}\:{W}_{{n},{p}−\mathrm{1}} \\ $$$$\Rightarrow{W}_{{n},{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \left({lnx}\right)^{{n}} \:−\frac{{n}}{{n}+\mathrm{1}}{W}_{{n},{n}−\mathrm{1}} \:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left({lnx}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}} \:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{{n}+\mathrm{1}}\:{W}_{{n}} \:\:\:\:…{be}\:{continued}…. \\ $$
Commented by Prithwish sen last updated on 27/Jun/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
$${you}\:{are}\:{welcome}\:{sir}. \\ $$