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)) ∫


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Question Number 6027 by FilupSmith last updated on 10/Jun/16

$x^{x} = e^{x \ln x}$ $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$ $∴ e^{x \ln x} = 1 + x \ln (x) + \frac{{(x \ln x)}^{2}}{2!} + . . .$ $e^{x \ln x} = \frac{{(x \ln x)}^{0}}{0!} + \frac{{(x \ln x)}^{1}}{1!} + \frac{{(x \ln x)}^{2}}{2!} + . . .$ $x^{x} = e^{x \ln x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n} {(\ln x)}^{n}}{n!}$ $Question :$ $\int_{0}^{1} x^{x} d x = \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} \frac{x^{n} {(\ln x)}^{n}}{n!} d x$ $by substituting :$ $x = \exp (- \frac{u}{n + 1}), 0 < u < \infty$ $show that :$ $\int_{0}^{1} x^{n} {(\ln x)}^{n} = {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int_{0}^{\infty} u^{n} e^{- u} d u$
Commented by FilupSmith last updated on 10/Jun/16

$\int_{0}^{1} x^{x} d x = \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} \frac{x^{n} {(\ln x)}^{n}}{n!} d x$ $x = e^{- \frac{u}{n + 1}} \Rightarrow u = - (n + 1) \ln (x)$ $d x = - \frac{1}{n + 1} e^{- \frac{u}{n + 1}}$ $x = 0 \Rightarrow u = \infty$ $x = 1 \Rightarrow u = 0$ $∴ \int_{0}^{1} x^{n} \ln {(x)}^{n} = \int_{\infty}^{0} e^{- \frac{u n}{n + 1}} \ln {(e^{- \frac{u}{n + 1}})}^{n} (- \frac{1}{n + 1} e^{- \frac{u}{n + 1}}) d u$ $= - \int_{0}^{\infty} e^{- (\frac{u n}{n + 1} + \frac{u}{n + 1})} \frac{1}{n + 1} (- \frac{u^{n}}{{(n + 1)}^{n}}) d u$ $= - {(n + 1)}^{- (n + 1)} \int_{0}^{\infty} e^{- u} u^{n} d u$ $My answer is missing {(- 1)}^{n} . Why ? ? ?$
Commented by FilupSmith last updated on 10/Jun/16

$on line 6 :$ $\int_{0}^{1} x^{n} \ln {(x)}^{n} = \int_{\infty}^{0} e^{- \frac{u n}{n + 1}} \ln {(e^{- \frac{u}{n + 1}})}^{n} (- \frac{1}{n + 1} e^{- \frac{u}{n + 1}}) d u$ $= - \int_{0}^{\infty} e^{- \frac{u n}{n + 1}} (\frac{{(- 1)}^{n} u^{n}}{{(n + 1)}^{n}}) (- \frac{1}{n + 1} e^{- \frac{u}{n + 1}}) d u$ $= {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int_{0}^{\infty} e^{- u} u^{n} d u$ $= {(- 1)}^{n} {(n + 1)}^{- (n + 1)} n!$ $\int_{0}^{1} x^{x} d x = \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{x^{n} \ln {(x)}^{n}}{n!} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)} n!$ $= \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} {(n + 1)}^{- (n + 1)}$ $∴ \int_{0}^{1} x^{x} d x = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} {(n + 1)}^{- (n + 1)}$
Commented by FilupSmith last updated on 10/Jun/16

$can we evaluate the RHS ?$
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