x^x =e^(xln x)
e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+...
∴ e^(xln x) = 1+xln(x)+(((xln x)^2 )/(2!))+...
e^(xln x) = (((xln x)^0 )/(0!))+(((xln x)^1 )/(1!))+(((xln x)^2 )/(2!))+...
x^x =e^(xln x) =Σ_(n=0) ^∞ ((x^n (ln x)^n )/(n!))
Question:
∫_0 ^( 1) x^x dx=Σ_(i=0) ^∞ ∫_0 ^1 ((x^n (ln x)^n )/(n!))dx
by substituting:
x=exp(−(u/(n+1))), 0<u<∞
show that:
∫_0 ^( 1) x^n (ln x)^n =(−1)^n (n+1)^(−(n+1)) ∫_0 ^( ∞) u^n e^(−u) du
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