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Question Number 7637 by Tawakalitu. last updated on 07/Sep/16
∫x^x dx
$$\int{x}^{{x}} \:{dx} \\ $$
Answered by FilupSmith last updated on 07/Sep/16
x^x =e^(xln(x)) =Σ_(n=0) ^∞ ((x^n ln(x)^n )/(n!)) ( ∵e^x =Σ_(n=0) ^∞ (x^n /(n!)) ) =Σ_(n=0) ^∞ (1/(n!))∫x^n ln(x)^n dx let x=e^(−(u/(n+1))) u=−(n+1)ln(x) dx=−(1/(n+1))e^(−(u/(n+1))) du dx=−(1/(n+1))e^(−(u/(n+1))) du ∫_a ^( b) x^x dx=Σ_(n=0) ^∞ (1/(n!))∫_( −(n+1)ln(a)) ^( −(n+1)ln(b)) (e^(−(u/(n+1))) )^n ln(e^(−(u/(n+1))) )^n (−(1/(n+1))e^(−(u/(n+1))) )du β=−(n+1)ln(b), α=−(n+1)ln(a) ∫_a ^( b) x^x dx=Σ_(n=0) ^∞ (1/(n!))∫_( α) ^( β) e^(−(((un)/(n+1))+(u/(n+1)))) (−(u/(n+1)))^n (−(1/(n+1)))du ∫_a ^( b) x^x dx=Σ_(n=0) ^∞ −(1/(n!))∫_( α) ^( β) e^(−(u/(n+1))(n+1)) (−1)^n (u^n /((n+1)^n ))(−(1/(n+1)))du ∫_a ^( b) x^x dx=Σ_(n=0) ^∞ −(1/(n!))∫_( α) ^( β) e^(−u) (−1)^n (u^n /((n+1)^(n+1) ))du ∫_a ^( b) x^x dx=Σ_(n=0) ^∞ (((−(−1)^n )/((n+1)^(n+1) n!))∫_( α) ^( β) e^(−u) u^n du)
$${x}^{{x}} ={e}^{{x}\mathrm{ln}\left({x}\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!} \\ $$$$\left(\:\:\because{e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}\:\:\right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} {dx} \\ $$$$\mathrm{let}\:{x}={e}^{−\frac{{u}}{{n}+\mathrm{1}}} \\ $$$${u}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$${dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} {du} \\ $$$${dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} {du} \\ $$$$ \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\:−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({a}\right)} ^{\:−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({b}\right)} \left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \mathrm{ln}\left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$\beta=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({b}\right),\:\:\:\:\alpha=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({a}\right) \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−\left(\frac{{un}}{{n}+\mathrm{1}}+\frac{{u}}{{n}+\mathrm{1}}\right)} \left(−\frac{{u}}{{n}+\mathrm{1}}\right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−\frac{{u}}{{n}+\mathrm{1}}\left({n}+\mathrm{1}\right)} \left(−\mathrm{1}\right)^{{n}} \frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−{u}} \left(−\mathrm{1}\right)^{{n}} \frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{−\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$ \\ $$
Commented by FilupSmith last updated on 07/Sep/16
for ∫_0 ^( 1) x^x dx, β=−(n+1)ln(1)=0 α=−(n+1)ln(0)=∞ ∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ (((−(−1)^n )/((n+1)^(n+1) n!))∫_( ∞) ^( 0) e^(−u) u^n du) ∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ ((((−1)^n )/((n+1)^(n+1) n!))∫_( 0) ^( ∞) e^(−u) u^n du) n!=∫_( 0) ^( ∞) e^(−u) u^n du ∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ ((((−1)^n )/((n+1)^(n+1) n!))n!) ∴∫_0 ^( 1) x^x dx=Σ_(n=0) ^∞ (−1)^n (n+1)^(−(n+1)) (≈0.783431)
$$\mathrm{for}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx},\:\:\beta=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{0}\right)=\infty \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{−\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\infty} ^{\:\mathrm{0}} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$${n}!=\int_{\:\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}{n}!\right) \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$$$\left(\approx\mathrm{0}.\mathrm{783431}\right) \\ $$
Commented by Tawakalitu. last updated on 07/Sep/16
Thanks so much, i really appreciate
$${Thanks}\:{so}\:{much},\:{i}\:{really}\:{appreciate} \\ $$
Commented by Tawakalitu. last updated on 07/Sep/16
Thanks sir
$${Thanks}\:{sir} \\ $$

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