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Question Number 6027 by FilupSmith last updated on 10/Jun/16
x^x =e^(xln x) e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+... ∴ e^(xln x) = 1+xln(x)+(((xln x)^2 )/(2!))+... e^(xln x) = (((xln x)^0 )/(0!))+(((xln x)^1 )/(1!))+(((xln x)^2 )/(2!))+... x^x =e^(xln x) =Σ_(n=0) ^∞ ((x^n (ln x)^n )/(n!)) Question: ∫_0 ^( 1) x^x dx=Σ_(i=0) ^∞ ∫_0 ^1 ((x^n (ln x)^n )/(n!))dx by substituting: x=exp(−(u/(n+1))), 0<u<∞ show that: ∫_0 ^( 1) x^n (ln x)^n =(−1)^n (n+1)^(−(n+1)) ∫_0 ^( ∞) u^n e^(−u) du
$${x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\therefore\:{e}^{{x}\mathrm{ln}\:{x}} \:=\:\mathrm{1}+{x}\mathrm{ln}\left({x}\right)+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$${e}^{{x}\mathrm{ln}\:{x}} \:=\:\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{0}} }{\mathrm{0}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{1}} }{\mathrm{1}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$${x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \left(\mathrm{ln}\:{x}\right)^{{n}} }{{n}!} \\ $$$$\mathrm{Question}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}} \left(\mathrm{ln}\:{x}\right)^{{n}} }{{n}!}{dx} \\ $$$$\mathrm{by}\:\mathrm{substituting}: \\ $$$${x}=\mathrm{exp}\left(−\frac{{u}}{{n}+\mathrm{1}}\right),\:\:\:\:\:\mathrm{0}<{u}<\infty \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} \left(\mathrm{ln}\:{x}\right)^{{n}} =\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{\mathrm{0}} ^{\:\infty} {u}^{{n}} {e}^{−{u}} {du} \\ $$
Commented by FilupSmith last updated on 10/Jun/16
∫_0 ^( 1) x^x dx=Σ_(i=0) ^∞ ∫_0 ^1 ((x^n (ln x)^n )/(n!))dx x=e^(− (u/(n+1))) ⇒ u=−(n+1)ln(x) dx=−(1/(n+1))e^(− (u/(n+1))) x=0 ⇒ u=∞ x=1 ⇒ u=0 ∴∫_0 ^( 1) x^n ln(x)^n = ∫_∞ ^( 0) e^(−((un)/(n+1))) ln(e^(−(u/(n+1))) )^n (−(1/(n+1))e^(− (u/(n+1))) )du =−∫_0 ^( ∞) e^(−(((un)/(n+1))+(u/(n+1)))) (1/(n+1))(−(u^n /((n+1)^n )))du =−(n+1)^(−(n+1)) ∫_0 ^( ∞) e^(−u) u^n du My answer is missing (−1)^n . Why???
$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}} \left(\mathrm{ln}\:{x}\right)^{{n}} }{{n}!}{dx} \\ $$$${x}={e}^{−\:\frac{{u}}{{n}+\mathrm{1}}} \:\:\Rightarrow\:\:{u}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$${dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\:\frac{{u}}{{n}+\mathrm{1}}} \\ $$$${x}=\mathrm{0}\:\Rightarrow\:{u}=\infty \\ $$$${x}=\mathrm{1}\:\Rightarrow\:{u}=\mathrm{0} \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} \:=\:\int_{\infty} ^{\:\mathrm{0}} {e}^{−\frac{{un}}{{n}+\mathrm{1}}} \mathrm{ln}\left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\:\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$=−\int_{\mathrm{0}} ^{\:\infty} {e}^{−\left(\frac{{un}}{{n}+\mathrm{1}}+\frac{{u}}{{n}+\mathrm{1}}\right)} \frac{\mathrm{1}}{{n}+\mathrm{1}}\left(−\frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\right){du} \\ $$$$=−\left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du} \\ $$$$\mathrm{My}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{missing}\:\left(−\mathrm{1}\right)^{{n}} .\:\mathrm{Why}??? \\ $$
Commented by FilupSmith last updated on 10/Jun/16
on line 6: ∫_0 ^( 1) x^n ln(x)^n = ∫_∞ ^( 0) e^(−((un)/(n+1))) ln(e^(−(u/(n+1))) )^n (−(1/(n+1))e^(− (u/(n+1))) )du =−∫_0 ^( ∞) e^(−((un)/(n+1))) ((((−1)^n u^n )/((n+1)^n )))(−(1/(n+1))e^(−(u/(n+1))) )du =(−1)^n (n+1)^(−(n+1)) ∫_0 ^( ∞) e^(−u) u^n du =(−1)^n (n+1)^(−(n+1)) n! ∫_0 ^( 1) x^x dx=Σ_(n=1) ^∞ ∫_0 ^( 1) ((x^n ln(x)^n )/(n!))dx =Σ_(n=1) ^∞ (1/(n!))(−1)^n (n+1)^(−(n+1)) n! =Σ_(n=1) ^∞ (−1)^n (n+1)^(−(n+1)) ∴ ∫_0 ^( 1) x^x dx=Σ_(n=1) ^∞ (−1)^n (n+1)^(−(n+1))
$$\mathrm{on}\:\mathrm{line}\:\mathrm{6}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} \:=\:\int_{\infty} ^{\:\mathrm{0}} {e}^{−\frac{{un}}{{n}+\mathrm{1}}} \mathrm{ln}\left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\:\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$=−\int_{\mathrm{0}} ^{\:\infty} {e}^{−\frac{{un}}{{n}+\mathrm{1}}} \left(\frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\right)\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} {n}! \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} {n}! \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$$$ \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$
Commented by FilupSmith last updated on 10/Jun/16
can we evaluate the RHS?
$$\mathrm{can}\:\mathrm{we}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{RHS}? \\ $$

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