TITLE: Derivative of $\int_{0}^{x} \sin(1/t) dt$ at $x= 0$
QUESTION [5 upvotes]: I've been trying to figure out how to evaluate 
$$
\frac{d}{dx}\int_{0}^{x} \sin(1/t) dt
$$ at $x = 0$. I know that the integrand is undefined at $x = 0,$ but is there any way to "extend" the derivative to the point? Or is it not differntiable there - and if so, why?

REPLY [2 votes]: Use the function $g(x) =x^{2}\cos(1/x),g(0)=0$ so that $g$ is differentiable with $$g'(x) =2x\cos(1/x)+\sin(1/x),g'(0)=0$$ and hence upon integrating we get $$\frac{1}{x}\int_{0}^{x}\sin(1/t)\,dt=\frac{g(x)}{x}-\frac{2}{x}\int_{0}^{x}t\cos(1/t)\,dt$$ Taking limits as $x\to 0$ we can see that the RHS tends to $0$ so the desired derivative is $0$.