Integral of xx {x}^{x}

Prove that ∫01xxdx=∑n=1∞(−1)n−1nn.\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=(eln⁡x)x=exln⁡x.{x}^{x} = {\left({e}^{\ln {x}} \right)}^{x} = {e}^{x \ln {x}}.

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

Thus ∫01xxdx=∑n=0∞∫01xn(ln⁡x)nn!=∑n=0∞1n!∫01xn(ln⁡x)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=(ln⁡x)nu = {\left(\ln {x} \right)}^{n} , dv=xndxdv = {x}^{n} dx , du=n(ln⁡x)n−1xdxdu = \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

lim⁡a→0∫a1xn(ln⁡x)ndx=lim⁡a→0[xn+1n+1(ln⁡x)n]a1−lim⁡a→0∫a1nn+1xn(ln⁡x)n−1dx\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 lim⁡a→0∫a1xn(ln⁡x)ndx=−∫01nn+1xn(ln⁡x)n−1dx=(−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)n−1nn.\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } } .

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#Calculus

Note by Steven Zheng
6 years, 6 months ago

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