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Question-138159




Question Number 138159 by BHOOPENDRA last updated on 10/Apr/21
Answered by Dwaipayan Shikari last updated on 10/Apr/21
∫_0 ^1 x^(−x) dx=ℵ  =∫_0 ^1 e^(−xlog(x)) dx=Σ_(n=0) ^∞ (((−1)^n )/(n!))∫_0 ^1 x^n log^n (x)dx=ℵ  ∫_0 ^1 x^n dx=(1/(n+1))⇒∫_0 ^1 x^n log^m (x)dx=(∂^m /∂n^m )((1/(n+1)))=((m!(−1)^m )/((n+1)^(m+1) ))  But here m=n   so  ℵ=Σ_(n=0) ^∞ (((−1)^n )/(n!)).(((−1)^n n!)/((n+1)^(n+1) ))=Σ_(n=1) ^∞ (1/n^n )
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−{x}} {dx}=\aleph \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{xlog}\left({x}\right)} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {log}^{{n}} \left({x}\right){dx}=\aleph \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx}=\frac{\mathrm{1}}{{n}+\mathrm{1}}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {log}^{{m}} \left({x}\right){dx}=\frac{\partial^{{m}} }{\partial{n}^{{m}} }\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\frac{{m}!\left(−\mathrm{1}\right)^{{m}} }{\left({n}+\mathrm{1}\right)^{{m}+\mathrm{1}} } \\ $$$${But}\:{here}\:{m}={n} \\ $$$$\:{so}\:\:\aleph=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}.\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{n}} } \\ $$$$ \\ $$

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