Integral of xx {x}^{x}

Prove that 01xxdx=n=1(1)n1nn.\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } } .

Solution

For convenience (as you will see later), let xx=(elnx)x=exlnx.{x}^{x} = {\left({e}^{\ln {x}} \right)}^{x} = {e}^{x \ln {x}}.

By the series expansion of ex{e}^{x}: exlnx=n=0(xlnx)nn!.{e}^{x \ln {x}} = \sum _{ n=0 }^{ \infty }{ \frac { { \left( x \ln{x} \right) }^{ n } }{ n! } } .

Thus 01xxdx=n=001xn(lnx)nn!=n=01n!01xn(lnx)ndx.\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } }=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } .

Let u=(lnx)nu = {\left(\ln {x} \right)}^{n} , dv=xndxdv = {x}^{n} dx , du=n(lnx)n1xdxdu = \frac{{n \left(\ln {x} \right)}^{n-1}}{x} dx and v=xn+1n+1v=\frac{{x}^{n+1}}{n+1}, then using integration by parts, we arrive at

lima0a1xn(lnx)ndx=lima0[xn+1n+1(lnx)n]a1lima0a1nn+1xn(lnx)n1dx\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =\lim _{ a\rightarrow 0 }{ { \left[ \frac { { x }^{ n+1 } }{ n+1 } { \left( \ln { x } \right) }^{ n } \right] }_{ a }^{ 1 } } -\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 } } dx

which becomes lima0a1xn(lnx)ndx=01nn+1xn(lnx)n1dx=(1)nn!(n+1)n+1.\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =-\int _{ 0 }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 }dx = \frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}.

Therefore, 01xxdx=n=1(1)n1nn.\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } } .

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Note by Steven Zheng
4 years, 10 months ago

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