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Question Number 6118 by sanusihammed last updated on 15/Jun/16
∫x^x dx
$$\int{x}^{{x}} \:{dx} \\ $$
Commented by FilupSmith last updated on 15/Jun/16
ANSWER: x^x =e^(xln(x)) e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+... ∴e^(xln(x)) =1+(xln(x))^1 +(((xln(x))^2 )/(2!))+... e^(xlnx) =Σ_(n=0) ^∞ ((x^n ln(x)^n )/(n!)) ∴∫x^x dx=∫ Σ_(n=0) ^∞ ((x^n ln(x)^n )/(n!))dx =Σ_(n=0) ^∞ ∫ ((x^n ln(x)^n )/(n!))dx =Σ_(n=0) ^∞ (1/(n!))∫ x^n ln(x)^n dx let x=e^(−(u/(n+1))) ∴dx=−(1/(n+1))e^(−(u/(n+1))) du =Σ_(n=0) ^∞ (1/(n!))∫ e^(−((un)/(n+1))) ln(e^(−(u/(n+1))) )^n (−(1/(n+1))e^(−(u/(n+1))) )du =Σ_(n=0) ^∞ (1/(n!))∫ e^(−((un)/(n+1))) ((((−u)^n )/((n+1)^n )))(−(1/(n+1))e^(−(u/(n+1))) )du =Σ_(n=0) ^∞ −(1/(n!))∫ e^(−(((un)/(n+1))+(u/(n+1)))) ((((−1)^n (u)^n )/((n+1)^n )))(1/(n+1))du ((un)/(n+1))+(u/(n+1))=((un+u)/(n+1))=((u(n+1))/((n+1)))=u =Σ_(n=0) ^∞ −(((−1)^n )/(n!))∫ e^(−u) (((u)^n )/((n+1)^(n+1) ))du ∴∫x^x dx=Σ_(n=0) ^∞ (−(1/(n!))(−1)^n (n+1)^(−(n+1)) ∫ e^(−u) u^n du) from Q6042 ∫e^(−u) u^n du=Cn!−e^(−u) n!(Σ_(r=0) ^n (u^(n−r) /((n−r)!))), C=constant
$$\mathrm{ANSWER}: \\ $$$${x}^{{x}} ={e}^{{x}\mathrm{ln}\left({x}\right)} \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\therefore{e}^{{x}\mathrm{ln}\left({x}\right)} =\mathrm{1}+\left({x}\mathrm{ln}\left({x}\right)\right)^{\mathrm{1}} +\frac{\left({x}\mathrm{ln}\left({x}\right)\right)^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$${e}^{{x}\mathrm{ln}{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!} \\ $$$$ \\ $$$$\therefore\int{x}^{{x}} {dx}=\int\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int\:\frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\:{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} {dx} \\ $$$$ \\ $$$$\mathrm{let}\:{x}={e}^{−\frac{{u}}{{n}+\mathrm{1}}} \\ $$$$\therefore{dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} {du} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\:{e}^{−\frac{{un}}{{n}+\mathrm{1}}} \mathrm{ln}\left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\:{e}^{−\frac{{un}}{{n}+\mathrm{1}}} \left(\frac{\left(−{u}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\right)\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{n}!}\int\:{e}^{−\left(\frac{{un}}{{n}+\mathrm{1}}+\frac{{u}}{{n}+\mathrm{1}}\right)} \left(\frac{\left(−\mathrm{1}\right)^{{n}} \left({u}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\right)\frac{\mathrm{1}}{{n}+\mathrm{1}}{du} \\ $$$$\frac{{un}}{{n}+\mathrm{1}}+\frac{{u}}{{n}+\mathrm{1}}=\frac{{un}+{u}}{{n}+\mathrm{1}}=\frac{{u}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)}={u} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int\:{e}^{−{u}} \frac{\left({u}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{du} \\ $$$$ \\ $$$$\therefore\int{x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int\:{e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$${from}\:{Q}\mathrm{6042} \\ $$$$\int{e}^{−{u}} {u}^{{n}} {du}={Cn}!−{e}^{−{u}} {n}!\left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right),\:{C}=\mathrm{constant} \\ $$
Commented by FilupSmith last updated on 15/Jun/16
an interesting/special result: ∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ (−(1/(n!))(−1)^n (n+1)^(−(n+1)) ∫_a ^( b) e^(−u) u^n du) x=exp(−(u/(n+1))) ⇒u=−(n+1)ln(x) a=−(n+1)ln(0) ∴a=−(−∞)=∞ b=−(n+1)ln(1) ∴b=0 ∴∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ (−(1/(n!))(−1)^n (n+1)^(−(n+1)) ∫_∞ ^^0 e^(−u) u^n du) =Σ_(n=0) ^∞ (+(1/(n!))(−1)^n (n+1)^(−(n+1)) ∫_0 ^^∞ e^(−u) u^n du) ∫_0 ^^∞ e^(−u) u^n du=n! =Σ_(n=0) ^∞ (((n!)/(n!))(−1)^n (n+1)^(−(n+1)) ) ∴∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ (−1)^n (n+1)^(−(n+1))
$${an}\:{interesting}/{special}\:{result}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{{a}} ^{\:{b}} \:{e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$${x}=\mathrm{exp}\left(−\frac{{u}}{{n}+\mathrm{1}}\right) \\ $$$$\Rightarrow{u}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$$\: \\ $$$${a}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{0}\right) \\ $$$$\therefore{a}=−\left(−\infty\right)=\infty \\ $$$$\: \\ $$$${b}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}\right) \\ $$$$\therefore{b}=\mathrm{0} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{\infty} ^{\:^{\mathrm{0}} } \:{e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(+\frac{\mathrm{1}}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \int_{\mathrm{0}} ^{\:^{\infty} } \:{e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$\int_{\mathrm{0}} ^{\:^{\infty} } \:{e}^{−{u}} {u}^{{n}} {du}={n}! \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{n}!}{{n}!}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \right) \\ $$$$\: \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$
Commented by sanusihammed last updated on 15/Jun/16
Thanks so much.
$${Thanks}\:{so}\:{much}. \\ $$

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