裴礼文 多元积分学 919页 练习7.2.25 解答

castelu

练习7.2.25：
　　计算积分
$$\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy$$

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解法1：
　　一方面
$$\begin{eqnarray*}
\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy&=&\int_0^{\pi}dx\int_x^{\pi}\ln\sin(y-x)dy\\
&=&\int_0^{\pi}dx\int_0^{\pi-x}\ln\sin udu
\end{eqnarray*}$$
　　另一方面
$$\begin{eqnarray*}
\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy&=&\int_0^{\pi}dy\int_0^y\ln\sin(y-x)dx\\
&=&\int_0^{\pi}dx\int_0^x\ln\sin(x-y)dy\\
&=&\int_0^{\pi}dx\int_0^x\ln\sin udu\\
&=&\int_0^{\pi}dx\int_{\pi-x}^{\pi}\ln\sin tdt\\
&=&\int_0^{\pi}dx\int_{\pi-x}^{\pi}\ln\sin udu
\end{eqnarray*}$$
　　两式相加
$$\begin{eqnarray*}
\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy&=&\frac{1}{2}\int_0^{\pi}dx\int_0^{\pi}\ln\sin udu\\
&=&\pi\int_0^{\frac{\pi}{2}}\ln\sin udu
\end{eqnarray*}$$
　　考虑积分
$$\int_0^{\frac{\pi}{2}}\ln\sin udu$$
　　易知
$$\int_0^{\frac{\pi}{2}}\ln\sin udu=\int_0^{\frac{\pi}{2}}\ln\cos udu$$
　　两式相加
$$\begin{eqnarray*}
\int_0^{\frac{\pi}{2}}\ln\sin udu&=&\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\frac{1}{2}\sin 2udu\\
&=&\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\ln\sin udu-\frac{\pi}{2}\ln2\right)
\end{eqnarray*}$$
　　故
$$\int_0^{\frac{\pi}{2}}\ln\sin udu=-\frac{\pi}{2}\ln2$$
　　那么
$$\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy=-\frac{\pi^2}{2}\ln2$$

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解法2：
　　由对称性
$$\begin{eqnarray*}
\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy&=&\frac{1}{2}\iint\limits_{[0,\pi] \times [0,\pi]}\ln |\sin(x-y)|dxdy\\
&=&\frac{1}{2}\int_0^{\pi}dx\int_0^{\pi}\ln |\sin(x-y)|dy\\
&=&\frac{1}{2}\int_0^{\pi}dx\int_{-x}^{\pi-x}\ln |\sin u|du\\
&=&\frac{1}{2}\int_0^{\pi}dx\int_0^{\pi}\ln |\sin u|du\\
&=&\frac{\pi}{2}\int_0^{\pi}\ln\sin udu\\
&=&\pi\int_0^{\frac{\pi}{2}}\ln\sin udu
\end{eqnarray*}$$
　　根据解法1的讨论
$$\iint\limits_{0 \le x \le y \le \pi}\ln |\sin(x-y)|dxdy=-\frac{\pi^2}{2}\ln2$$