The antiderivative of $\sin(1/x)$ How to prove that the function $f(x)=\sin\frac{1}{x}$ for $x\neq 0,f(0)=0$ has an antiderivative? This means $F(x)=\int^{x}_{0}\sin(1/t)dt$ has derivative $0$ at $x=0$, but I have no idea how to prove it.
 A: I just blogged another solution to this question in my blog. The idea is to consider the following function
$$G(x)=\begin{cases}x^2\cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$$
which is differentiable. Clearly,
$$G'(x)=\begin{cases} \sin \frac 1x + 2x \cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$$
Hence, $G' = f + h$ where
$$h(x)=\begin{cases} 2x \cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$$
Since $h$ is continuous, it has antiderivative $H$, thus giving us $f = (G-H)'$. In other words, $G-H$ is an antiderivative of $f$.
A: We can substitute $u = t^{-1}$ to get a more convenient expression:
$$\begin{align}
\left\lvert \frac{F(x) - F(0)}{x}\right\rvert
&= \frac{1}{\lvert x\rvert} \left\lvert \int_0^x \sin (t^{-1})\,dt\right\rvert \\
&= \frac{1}{\lvert x\rvert} \left\lvert \int_{1/\lvert x\rvert}^\infty \frac{\sin u}{u^2}\,du\right\rvert \tag{symmetry}\\
&= \frac{1}{\lvert x\rvert} \left\lvert \left[-\frac{\cos u}{u^2}\right]_{1/\lvert x\rvert}^\infty - 2 \int_{1/\lvert x\rvert}^\infty \frac{\cos u}{u^3}\,du\right\rvert\\
&= \frac{1}{\lvert x\rvert} \left\lvert \lvert x\rvert^2 \cos \frac{1}{\lvert x\rvert} - 2 \int_{1/\lvert x\rvert}^\infty \frac{\cos u}{u^3}\,du\right\rvert\\
&\leqslant \lvert x\rvert + \frac{1}{\lvert x\rvert} \int_{1/\lvert x\rvert}^\infty \frac{2}{u^3}\,du\\
&= 2\lvert x\rvert.
\end{align}$$
Thus $F'(0) = 0$.
