TITLE: Differentiablility of a function
QUESTION [7 upvotes]: Given: $f(x)=\cos(1/x)$ When $x \neq 0$ and zero at $x=0$, is:
$$
F(x)= \int^x_0f(t)\,dt
$$
differentiable at zero?
I believe this is differntiable, since intuitively we are shrinking the area we are integrating over so it should approach zero, but I could use a hand starting the proof.

REPLY [4 votes]: Here is a tedious (but elementary) approach:
We have ${F(h) \over h} = {1 \over h} \int_0^h \cos {1 \over x} dx$. Using the substitution $t={1 \over x}$ we get $\int_0^h \cos {1 \over x} dx =\int_{1 \over h}^\infty \cos t {1 \over t^2} dt$, and integrating by parts with $u = {1 \over t^2}, dv = \cos t \,dt$ gives
$\int_{1 \over h}^\infty \cos t {1 \over t^2} dt = {1 \over t^2} \sin t \big |_{1 \over h}^\infty+ \int_{1 \over h}^\infty\sin t {2 \over t^3} dt$.
Hence $|\int_0^h \cos {1 \over x} dx| \le |h^2 \sin {1 \over h}|+ \int_{1 \over h}^\infty {2 \over t^3} dt \le h^2+h^2$, from which we get
$|{F(h) \over h}| \le 2h$, and hence $F'(0) = 0$.