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

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