求解一道2015年北大直博招生题目

证明(本题30分):$\int_{0}^{1}\int_{0}^{1} \frac{1}{(xy)^{xy}}dxdy=\int_0^1\frac{1}{x^x}dx=\sum\limits_{n=1}^{+ \infty} \frac{1}{n^n}$.
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Eufisky - 熊熊

赞同来自: 代数龙

令$u = xy,v = x$,则$x=v,y=\frac uv$.由$0\leq x,y\leq 1$可知$0\leq u\leq v,0\leq v\leq 1$,
\begin{align*}
\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {\begin{array}{*{20}{c}}
0&1\\
{\frac{1}{v}}&{ - \frac{u}{{{v^2}}}}
\end{array}} \right| = - \frac{1}{v}\,,\end{align*}
那么有
\begin{align*}
&\int_0^1 {\int_0^1 {\frac{1}{{{{\left( {xy} \right)}^{xy}}}}dxdy} } = \int_0^1 {dv} \int_0^v {\frac{1}{{{u^u}v}}du} \\
&= \int_0^1 {du} \int_u^1 {\frac{1}{{{u^u}v}}dv} = \int_0^1 {\frac{{ - \ln u}}{{{u^u}}}du} \\
&= \int_0^1 {\frac{{ - \ln u - 1}}{{{u^u}}}du} + \int_0^1 {\frac{1}{{{u^u}}}du} \\
&= \left[ {\frac{1}{{{u^u}}}} \right]_0^1 + \int_0^1 {\frac{1}{{{u^u}}}du} = \int_0^1 {\frac{1}{{{x^x}}}dx}.
\end{align*}

\begin{align*}
&\int_0^1 {\frac{1}{{{x^x}}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} \\
&= \int_0^1 {\sum\limits_{n = 0}^{+\infty} {\frac{{{{\left( { - x\ln x} \right)}^n}}}{{n!}}} dx} = \sum\limits_{n = 0}^{+\infty} {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} }.
\end{align*}
令$t=-(n+1)\ln x$,有
\begin{align*}
\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} &= \frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}\int_0^{ + +\infty } {{t^n}{e^{ - t}}dt} \\
&= \frac{{\Gamma \left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^{n + 1}}}} = \frac{{n!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}.
\end{align*}
因此有

\begin{align*}
\int_0^1 {\frac{1}{{{x^x}}}dx} = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} } = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^n}}}} .\end{align*}

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