问题及解答
1
回忆三角函数的和差化积公式
\[
\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}.
\]
因此
\[
\sin(2n-1)x-\sin(2n-3)x=2\sin x\cos 2(n-1)x.
\]
于是, 有
\[
\begin{eqnarray}
\sin(2n-1)x-\sin(2n-3)x&=&2\sin x\cos 2(n-1)x,\\
\sin(2n-3)x-\sin(2n-5)x&=&2\sin x\cos 2(n-2)x,\\
\vdots &=& \vdots\\
\sin 3x-\sin x &=& 2\sin x\cos 2x.
\end{eqnarray}
\]
将以上 $n-1$ 个式子相加, 得到
\[
\sin(2n-1)x-\sin x=2\sin x\cdot\bigl[\cos 2x+\cos 4x+\cdots+\cos 2(n-1)x\bigr],
\]
因此
\[
\frac{\sin(2n-1)x}{\sin x}=1+2\bigl[\cos 2x+\cos 4x+\cdots+\cos 2(n-1)x\bigr].
\]
于是,
\[
\int_{0}^{\frac{\pi}{2}}\frac{\sin(2n-1)x}{\sin x}dx=\frac{\pi}{2}+2\sum_{k=1}^{n-1}\int_{0}^{\frac{\pi}{2}}\cos 2kx dx=\frac{\pi}{2}.\quad(n\geqslant 1).
\]
再利用等式 $\sin^2 nx-\sin^2(n-1)x=\sin(2n-1)x\sin x$ 可得
\[
\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{\sin^2 x}dx=\sum_{k=1}^{n}\int_{0}^{\frac{\pi}{2}}\frac{\sin(2k-1)x}{\sin x}dx=\frac{\pi}{2}n,\quad n\geqslant 1.
\]
事实上, 根据公式 $\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$, 得
\[
\begin{split}
\sin^2 nx-\sin^2(n-1)x&=\frac{1-\cos 2nx}{2}-\frac{1-\cos 2(n-1)x}{2}=\frac{1}{2}\cdot\Bigl[\cos(2n-2)x-\cos 2nx\Bigr]\\
&=-1\cdot\sin(2n-1)x\cdot\sin(-1)x\\
&=\sin(2n-1)x\cdot\sin x.
\end{split}
\]
于是有
\[
\begin{eqnarray}
\sin^2 nx-\sin^2(n-1)x&=&\sin(2n-1)x\cdot\sin x,\\
\sin^2 (n-1)x-\sin^2(n-2)x&=&\sin(2n-3)x\cdot\sin x,\\
\vdots &=& \vdots\\
\sin^2 2x-\sin^2 x&=&\sin 3x\cdot\sin x,\\
\end{eqnarray}
\]
将上面 $n-1$ 个式子相加, 得
\[
\sin^2 nx-\sin^2 x=\sin x\Bigl[\sin 3x+\sin 5x+\cdots+\sin (2n-1)x\Bigr]=\sin x\cdot\sum_{k=2}^{n}\sin (2k-1)x.
\]
从而
\[
\frac{\sin^2 nx}{\sin^2 x}=1+\frac{\sin 3x+\sin 5x+\cdots+\sin (2n-1)x}{\sin x}=\frac{\sum_{k=1}^{n}\sin (2k-1)x}{\sin x}.
\]
于是
\[
\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{\sin^2 x}dx=\sum_{k=1}^{n}\int_{0}^{\frac{\pi}{2}}\frac{\sin (2k-1)x}{\sin x}dx=\frac{\pi}{2}n.
\]
最后我们将比较下面两个积分
\[
\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{x^2}dx,\quad\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{\sin^2 x}dx,
\]
在区间 $[0,\frac{\pi}{2}]$ 上, 利用不等式 $x-\frac{x^3}{3}\leqslant\sin x\leqslant x$, 可得
\[
\begin{split}
\frac{1}{n}\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{x^2}dx&\leqslant\frac{\pi}{2}=\frac{1}{n}\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 nx}{\sin^2 x}dx\\
&\leqslant\frac{1}{n}\cdot(1-\frac{2}{3}\delta^2)^{-1}\int_{0}^{\delta}\frac{\sin^2 nx}{x^2}dx+\frac{1}{n}\cdot(1-\frac{1}{6}\pi^2)^{-1}\int_{\delta}^{\frac{\pi}{2}}\frac{1}{x^2}dx.
\end{split}
\]
这里要注意, 当 $0<x\leqslant\delta$ 时,
\[
\frac{1}{\sin^2 x}\leqslant\frac{1}{x^2(1-\frac{x^2}{3})^2}<\frac{1}{x^2}\cdot\frac{1}{1-\frac{2}{3}\delta^2}.
\]
当 $\delta<x<\frac{\pi}{2}$ 时,
\[
\frac{\sin^2 nx}{\sin^2 x}\leqslant\frac{1}{\sin^2 x}\leqslant\frac{1}{x^2(1-\frac{x^2}{3})^2}\leqslant\frac{1}{x^2}\cdot\frac{1}{1-\frac{\pi^2}{6}}.
\]
作变量代换 $x=\frac{t}{n}$, 得
\[
\int_{0}^{\frac{n}{2}\pi}\frac{\sin^2 t}{t^2}dt\leqslant\frac{\pi}{2}\leqslant (1-\frac{2}{3}\delta^2)^{-1}\int_{0}^{n\delta}\frac{\sin^2 t}{t^2}dt+(1-\frac{1}{6}\pi^2)^{-1}\cdot\int_{n\delta}^{\frac{\pi}{2}n}\frac{1}{t^2}dt.
\]
并令 $n\rightarrow+\infty$, 得
\[
\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}dx\leqslant\frac{\pi}{2}\leqslant (1-\frac{2}{3}\delta^2)^{-1}\cdot\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}dx.
\]
再令 $\delta\rightarrow 0^{+}$, 即得
\[
\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}dx=\frac{\pi}{2}.
\]