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Question-120531




Question Number 120531 by A8;15: last updated on 01/Nov/20
Commented by Dwaipayan Shikari last updated on 01/Nov/20
∫e^x^x  dx=∫Σ_(n=0) ^∞ (((x^x )^n )/(n!))dx  =Σ_(n=0) ^∞ (1/(n!))∫x^(nx) dx  =Σ_(n=0) ^∞ (1/(n!))∫(e^(xlogx) )^n dx  =Σ_(n=0) ^∞ (1/(n!))∫e^(nxlogx) dx  =Σ_(n=0) ^∞ (1/(n!))∫Σ_(n=0) ^∞ (((nxlogx)^n )/(n!))dx  =Σ_(n=0) ^∞ (1/(n!))Σ_(n=0) ^∞ (1/(n!))∫n^n x^n log^n xdx    ∫_0 ^1 x^x dx=Σ_(n=0) ^∞ (((−1)^n )/((n+1)^((n+1)) ))
$$\int{e}^{{x}^{{x}} } {dx}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({x}^{{x}} \right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{x}^{{nx}} {dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\left({e}^{{xlogx}} \right)^{{n}} {dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{e}^{{nxlogx}} {dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({nxlogx}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{n}^{{n}} {x}^{{n}} {log}^{{n}} {xdx} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\left({n}+\mathrm{1}\right)} } \\ $$
Answered by Lordose last updated on 01/Nov/20
I prefer you use series
$$\mathrm{I}\:\mathrm{prefer}\:\mathrm{you}\:\mathrm{use}\:\mathrm{series} \\ $$

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