Differentiability of a integral

Question

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.

By $\cos(1/u)^2$, do you mean $\cos^2(1/u)$, or $[\cos(1/u)]^2$? — Brian Tung, Jun 08 '16 at 17:55
@BrianTung I would say neither. It's the cosine of the quantity (1/u)^2 — zhw., Jun 08 '16 at 18:16
This reminds of a similar problem http://math.stackexchange.com/q/1784485/72031 — Paramanand Singh, Jun 08 '16 at 18:22
@zhw.: Yes, that's what I meant to type for my second choice. As it stands, my two choices are identical, of course. (Although it turns out that OP did indeed mean $\cos^2(1/u)$.) — Brian Tung, Jun 08 '16 at 18:32
@ParamanandSingh a similar technique would work if I can find a function with derivative the integrated plus an continuous term, but I haven't found one — davik, Jun 08 '16 at 18:34
Having given it some more thought, I still think this is continuous but not differentiable at $0$ (retracted too quickly on GFauxPas's answer). Willing to be shown wrong, though. — Brian Tung, Jun 08 '16 at 18:34

zhw. · Accepted Answer · 2016-06-09T15:26:18.863

4

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.

edited Jun 09 '16 at 15:26

answered Jun 08 '16 at 18:41

zhw.

105,693

Sorry I think you misunderstood, the cosine is squared. – davik Jun 08 '16 at 18:53
I didn't misunderstand, you changed the problem. I think things will be easier with the change. – zhw. Jun 08 '16 at 19:00
I edited my answer. It is easier now. – zhw. Jun 08 '16 at 19:09
Ok I think I agree that the limit exists for your sequence, but I don't see how to finish, do you see an easy way? – davik Jun 08 '16 at 19:15
Let $x_n = 1/(2n\pi).$ If $x_{n+1}\le x \le x_n,$ then
$$\frac{L(x_{n+1})}{x_n} \le \frac{L(x)}{x} \le \frac{L(x_{n})}{x_{n+1}}.$$

On the right we have $(L(x_n)/x_n) \cdot (x_n/x_{n+1}).$ Same idea on the left. The fact that $x_n/x_{n+1}\to 1$ gives it.
– zhw. Jun 08 '16 at 19:26
Thanks, I noticed that this generalizes to the integral of any periodic function evaluated at 1/u – davik Jun 08 '16 at 19:51
In fact to $u^{-p}$ as long as $p > 1$ – davik Jun 08 '16 at 20:02
Really nice argument. +1 – Paramanand Singh Jun 09 '16 at 06:01
@ParamanandSingh Thank you. – zhw. Jun 09 '16 at 15:26

Differentiability of a integral

1 Answers1

Linked