问题及解答

证明

\[\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx < \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx.\]

问题等价于证明

\[
\int_0^{\frac{\pi}{2}}\frac{\cos x-\sin x}{1+x^2}dx > 0.
\]

考虑函数

\[
f(x)=\int_0^{x}\frac{\cos t-\sin t}{1+t^2}dt.
\]

则

\[
f'(x)=\frac{\cos x-\sin x}{1+x^2}=\frac{\sqrt{2}}{1+x^2}\cdot\sin(\frac{\pi}{4}-x).
\]

易见

\[
f'(x)\begin{cases}
< 0, & x\in(0,\frac{\pi}{4}),\\
=0, & x=\frac{\pi}{4},\\
>0, & x\in(\frac{\pi}{4},\frac{\pi}{2}).
\end{cases}
\]

因此, $x=\frac{\pi}{4}$ 是 $f(x)$ 的最小值点. 而

\[
f(\frac{\pi}{4})=\int_0^{\frac{\pi}{4}}\frac{\cos t-\sin t}{1+t^2}dt=\sqrt{2}\int_0^{\frac{\pi}{4}}\frac{\sin(\frac{\pi}{4}-t)}{1+t^2}dt=\sqrt{2}\cdot\sin(\frac{\pi}{4}-t_0)\int_0^{\frac{\pi}{4}}\frac{1}{1+t^2}dt > 0.
\]

这里用到了中值定理, $t_0\in(0,\frac{\pi}{4})$.