Differentiability of a integral Define $$L(x)=\int_0^x \left(\sqrt{1+\cos^2(1/u)}\right)du$$, then is L right-differentiable at 0?
As far as I'm aware, l'hopitals rule doesn't apply here, but I still think it should be differentiable, because the oscillations speed up as you approach 0.
 A: Here's an outline of a proof that $L$ is right differentiable at $0$: Because $L(0) = 0,$ we need to show that $L(x)/x$ has a finite limit as $x\to 0^+.$ Let $u = 1/v.$ Then $L(x)/x$ equals
$$\frac{1}{x}\int_{1/x}^\infty \frac{\sqrt {1+\cos^2 v}}{v^2}\,dv.$$
Consider $x= 1/(2n\pi).$ We are then looking at
$$\tag 1  2n\pi\int_{2n\pi}^\infty \frac{\sqrt {1+\cos^2 v}}{v^2}\,dv.$$
Let $I = \int_0^{2\pi} \sqrt {1+\cos^2 t}\, dt.$ Think about what happens on intervals of length $2\pi$ to see $(1)$ is bounded above by
$$ 2n\pi\sum_{k=n}^{\infty} I\cdot (2\pi k)^{-2}.$$
Leaving out a bunch of constants, the crux of the matter is what happens with
$$ n \sum_{k=n}^{\infty} k^{-2}$$
as $n\to \infty.$ This has the limit of $1.$ (Approximate the sum with $\int_n^\infty t^{-2}\,dt.$)
The same argument can be made from below, replacing $(2\pi k)^{-2}$ by $(2\pi (k+1))^{-2}$ etc. You get the same limit.
OK, you still need to verify that taking $x= 1/(2n\pi)$ is enough to prove it. Modulo this, we've shown $L'(0)$ exists from the right.
