设 $a$ 为大于 1 的常数, $f(x)$ 是连续函数,证明
$$
\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x=\int_1^a f\left(x+\frac{a^2}{x}\right) \frac{1}{x} \mathrm{~d} x .
$$
【答案】 18. 【参考证明】设 $x=\sqrt{t}$ ,则有
$$
\begin{aligned}
&\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x=\frac{1}{2} \int_1^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t \\
&=\frac{1}{2}\left[\int_1^a f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t+\int_a^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t\right]
\end{aligned}
$$
又令 $t=\frac{a^2}{u}$ 时,有
$$
\begin{aligned}
&\int_a^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t=-\int_a^1 f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u \\
&=\int_1^a f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u
\end{aligned}
$$
所以
$$
\begin{aligned}
&\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x \\
&=\frac{1}{2}\left[\int_1^a f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t+\int_1^a f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u\right] \\
&=\int_1^a f\left(x+\frac{a^2}{x}\right) \frac{1}{x} \mathrm{~d} x .
\end{aligned}
$$


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