设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.
设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.
设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.
1
\[
\begin{split}
\int_0^1 I(x)dx&=\int_0^1\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dydx\\
&=\int_{\pi}^{2\pi}\biggl(\int_0^1\frac{y\sin(xy)}{y-\sin y}dx\biggr)dy\\
&=\int_{\pi}^{2\pi}\frac{1}{y-\sin y}\biggl(\int_0^1 y\sin(xy)dx\biggr)dy\\
&=\int_{\pi}^{2\pi}\frac{1}{y-\sin y}\biggl(\int_0^1 \sin(xy)d(xy)\biggr)dy\\
&=\int_{\pi}^{2\pi}\frac{1}{y-\sin y}\biggl(\int_0^y \sin tdt\biggr)dy\\
&=\int_{\pi}^{2\pi}\frac{1}{y-\sin y}[(-\cos t)\biggr|_{0}^{y}]dy\\
&=\int_{\pi}^{2\pi}\frac{1-\cos y}{y-\sin y}dy\\
&=\int_{\pi}^{2\pi}\frac{1}{y-\sin y}d(y-\sin y)\\
&\stackrel{u=y-\sin y}{=}\int_{\pi}^{2\pi}\frac{du}{u}\\
&=\ln(2\pi)-\ln\pi=\ln 2.
\end{split}
\]