解析入門Ⅰ 第Ⅳ章積分法 §５一変数函数の積分　大問5 問題の解法

間違いがあればコメントで教えていただければ幸いです。

$${\bf5)\quad}$$$${f}$$が連続のとき次の等式を証明せよ。

$${(\mathrm{i})\quad \displaystyle\int^{a^{2}}_{1}\!\!f\!\left(x+\dfrac{a^{2}}{x}\right)\!\dfrac{dx}{x}=2\!\int^{a}_{1}\!\!f\!\left(x^{2}+\dfrac{a^{2}}{x^{2}}\right)\!\dfrac{dx}{x}\quad(a\gt0)}$$

$$
\displaystyle ここでx=t^{2}と置換すると、\dfrac{dx}{dt}=2t,\def\arraystretch{1.3}\begin{array}{c:c}x&1\to a^{2}\\\hline t&1\to a^{   }\end{array}\\\begin{aligned}\therefore\int^{a^{2}}_{1}\!\!f\!\left(x+\dfrac{a^{2}}{x}\right)\!\dfrac{dx}{x}&=\int^{a}_{1}\!\!f\!\left(t^{2}+\dfrac{a^{2}}{t^{2}}\right)\!\dfrac{2t\,dt}{t^{2}}\\&=2\!\int^{a}_{1}\!\!f\!\left(t^{2}+\dfrac{a^{2}}{t^{2}}\right)\!\dfrac{dt}{t}\\&=2\!\int^{a}_{1}\!\!f\!\left(x^{2}+\dfrac{a^{2}}{x^{2}}\right)\!\dfrac{dx}{x}\end{aligned}
$$

$${(\mathrm{ii})\quad \displaystyle\int^{a}_{1}\!\!f\!\left(x+\dfrac{a^{2}}{x}\right)\!\dfrac{dx}{x}=\!\int^{a^{2}}_{a}\!\!f\!\left(x+\dfrac{a^{2}}{x}\right)\!\dfrac{dx}{x}\quad(a\gt0)}$$

$$
\displaystyle ここでx=\dfrac{a^{2}}{t}と置換すると、\dfrac{dx}{dt}=-\dfrac{a^{2}}{t^{2}},\def\arraystretch{1.3}\begin{array}{c:c}x&1\to a\\\hline t&a^{2}\!\to a\end{array}\\\begin{aligned}\therefore\int^{a}_{1}\!\!f\!\left(x+\dfrac{a^{2}}{x}\right)\!\dfrac{dx}{x}&=\int^{a}_{a^{2}}f\left(\dfrac{a^{2}}{t}+t\right)\dfrac{t}{a^{2}}\left(-\dfrac{a^{2}}{t^{2}}dt\right)\\&=-\int^{a}_{a^{2}}f\left(t+\dfrac{a^{2}}{t}\right)\dfrac{dt}{t}\\&=\int^{a^{2}}_{a}f\left(t+\dfrac{a^{2}}{t}\right)\dfrac{dt}{t}\\&=\int^{a^{2}}_{a}f\left(x+\dfrac{a^{2}}{x}\right)\dfrac{dx}{x}\end{aligned}
$$