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Question Number 10913 by Saham last updated on 01/Mar/17

∫_( 1) ^( 3)  x^x   dx

$$\int_{\:\mathrm{1}} ^{\:\mathrm{3}} \:\mathrm{x}^{\mathrm{x}} \:\:\mathrm{dx} \\ $$

Answered by FilupS last updated on 02/Mar/17

I=∫_1 ^( 3) x^x dx  I=∫_1 ^( 3) e^(xln(x)) dx  e^t =Σ_(n=0) ^∞ (t^n /(n!))  ∴ x^x =e^(xln(x)) =Σ_(n=0) ^∞ ((x^n ln^n (x))/(n!))  I=∫_1 ^( 3) Σ_(n=0) ^∞ ((x^n ln^n (x))/(n!))dx  I=Σ_(n=0) ^∞ ∫_1 ^( 3) ((x^n ln^n (x))/(n!))dx  I=Σ_(n=0) ^∞ (1/(n!))∫_1 ^( 3) x^n ln^n (x)dx

$${I}=\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}^{{x}} {dx} \\ $$$${I}=\int_{\mathrm{1}} ^{\:\mathrm{3}} {e}^{{x}\mathrm{ln}\left({x}\right)} {dx} \\ $$$${e}^{{t}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{n}} }{{n}!} \\ $$$$\therefore\:{x}^{{x}} ={e}^{{x}\mathrm{ln}\left({x}\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}^{{n}} \left({x}\right)}{{n}!} \\ $$$${I}=\int_{\mathrm{1}} ^{\:\mathrm{3}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}^{{n}} \left({x}\right)}{{n}!}{dx} \\ $$$${I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{1}} ^{\:\mathrm{3}} \frac{{x}^{{n}} \mathrm{ln}^{{n}} \left({x}\right)}{{n}!}{dx} \\ $$$${I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}^{{n}} \mathrm{ln}^{{n}} \left({x}\right){dx} \\ $$

Commented by Saham last updated on 02/Mar/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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