被积函数不连续 ¶

无法直接对变上限积分求导，先分部积分

\(f(x) = \displaystyle\int_0^x\sin\dfrac{1}{t}\mathrm{d}t,求f'(0)\)

\(f(x) = -\displaystyle\int_0^xt^2\mathrm{d}\left(\cos\dfrac{1}{t}\right) = -x^2\cos\dfrac{1}{x} + 2\displaystyle\int_0^xt\cos\dfrac{1}{t}\mathrm{d}t\)

\(f'(0) = \lim\limits_{x\to0}\dfrac{f(x) - f(0)}{x} = \lim\limits_{x\to0}\dfrac{ 2\displaystyle\int_0^xt\cos\dfrac{1}{t}\mathrm{d}t}{x} = \lim\limits_{x\to0}2x\cos\dfrac{1}{x} = 0\)