数分练习[3]

  证明:
$$\int_0^1\int_0^1(xy)^{xy}\,\mathrm dx\,\mathrm dy=\int_0^1y^y\,\mathrm dy.$$


  证明 令$\,xy=t$,那么\begin{align*}\int_0^1\int_0^1(xy)^{xy}\,\mathrm dx\,\mathrm dy&=\int_0^1\dfrac1y\int_0^yt^t\,\mathrm dt\,\mathrm dy\\&=\int_0^1\int_0^yt^t\,\mathrm dt\,\mathrm d\ln y\\&=\left.\ln y\cdot\int_0^yt^t\,\mathrm dt\right|_0^1-\int_0^1\ln y\,\mathrm d\int_0^yt^t\,\mathrm dt\\&=-\lim_{y\to0}\ln y\cdot\int_0^yt^t\,\mathrm dt-\int_0^1y^y\ln y\,\mathrm dy.\end{align*}熟知$$\lim_{y\to0}y\ln y=0,$$由洛必达法则又有$$\lim_{y\to0}\dfrac{\int_0^yt^t\,\mathrm dt}{y}=\lim_{y\to0}y^y=1,$$所以$$\lim_{y\to0}\ln y\cdot\int_0^yt^t\,\mathrm dt=\lim_{y\to0}y\ln y\cdot\lim_{y\to0}\dfrac{\int_0^yt^t\,\mathrm dt}{y}=0,$$因此$$\int_0^1\int_0^1(xy)^{xy}\,\mathrm dx\,\mathrm dy=-\int_0^1y^y\ln y\,\mathrm dy.$$注意到$$\left(y^y\right)’=y^y\ln y+y^y,$$所以$$\int_0^1\int_0^1(xy)^{xy}\,\mathrm dx\,\mathrm dy=\int_0^1\left(y^y-\left(y^y\right)’\right)\,\mathrm dy=\int_0^1y^y\,\mathrm dy.$$


   这题的结果十分漂亮!

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