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# Integral of $${x}^{x}$$

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

Solution

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

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

Thus $\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } }=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } .$

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

$\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =\lim _{ a\rightarrow 0 }{ { \left[ \frac { { x }^{ n+1 } }{ n+1 } { \left( \ln { x } \right) }^{ n } \right] }_{ a }^{ 1 } } -\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 } } dx$

which becomes $\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =-\int _{ 0 }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 }dx = \frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}.$

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

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Note by Steven Zheng
2 years, 9 months ago

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