# Differentiability of an integral accumulation function

Is $$H(x) = \int_0^x \left\lvert\sin\left(\frac{1}t\right)\right\rvert\,\mathrm dt$$ differentiable at $$x = 0$$?

I claim that $$H(x)$$ is differentiable at $$x=0.$$ Observe that \begin{align}H(-x) &= \displaystyle\int_0^{-x}|\sin(\frac{1}t)|dt=\displaystyle\int_0^x |\sin(-\frac{1}u)|(-1)du,\text{ where u = -t,}\\ &=-\displaystyle\int_0^x |\sin(\frac{1}t)|dt = -H(x),\end{align}

so $$H(x)$$ is odd. Also, $$H(0) = 0.$$ It suffices to evaluate $$\lim\limits_{x\to 0^+}\dfrac{H(x)}x,$$ since if $$\lim\limits_{x\to 0^+} H(x)$$ exists, it must equal $$-\lim\limits_{x\to 0^-}H(x),$$ which implies that $$\lim\limits_{x\to 0^+}\dfrac{H(x)}x = \lim\limits_{x\to 0^-}\dfrac{H(x)}x.$$

So assume $$x>0.$$ Since $$|\sin(\frac{1}t)|$$ is bounded and continuous on $$(0, x], H(x) = \displaystyle\int_0^x |\sin(\frac{1}t)|dt = \lim\limits_{u\to 0^+}\displaystyle\int_u^x |\sin(\frac{1}t)|dt=\lim\limits_{n\to\infty}\displaystyle\int_{1/((n+1)\pi)}^{1/(k_x\pi)}|\sin(\frac{1}t)|dt + \displaystyle\int_{1/(k_x\pi)}^x |\sin(\frac{1}t)|dt \\ = \displaystyle\sum_{k=k_x}^\infty \displaystyle\int_{1/((k+1)\pi)}^{1/(k\pi)}|\sin(\frac{1}t)|dt+\displaystyle\int_{1/(k_x\pi)}^x|\sin(\frac{1}t)|dt,$$

where $$\frac{1}{k_x\pi} \leq x \leq \frac{1}{(k_x-1)\pi}\Rightarrow k_x\pi \geq \frac{1}{x} \geq (k_x - 1)\pi \Rightarrow k_x = \lceil \frac{1}{x\pi} \rceil.$$

Now, observe that $$0 \leq |\dfrac{H(x)}x|\leq \dfrac{1}x \left|\displaystyle\sum_{k=k_x}^\infty \displaystyle\int_{1/((k+1)\pi)}^{1/(k\pi)}|\sin(\frac{1}t)|dt + \displaystyle\int_{1/(k_x\pi)}^x |\sin(1/t)|dt\right|\\ \leq \dfrac{1}x(\left|\displaystyle\sum_{k=k_x}^\infty \displaystyle\int_{1/((k+1)\pi)}^{1/(k\pi)}|\sin(\frac{1}t)|dt\right|+\left|\displaystyle\int_{1/(k_x\pi)}^x |\sin(1/t)|dt\right|)\leq \dfrac{1}x(\lim\limits_{n\to\infty} \dfrac{1}{k_x\pi} - \dfrac{1}{(n+1)\pi}+x-\dfrac{1}{k_x\pi})\leq 1,$$

however, here I am stuck. Also, $$\dfrac{H(x)}{x}$$ is not monotone, so I think I should use a different approach.

I know that for $$t\in [\dfrac{1}{n\pi+\frac{3\pi}4}, \dfrac{1}{n\pi+\frac\pi4}], |\sin(\dfrac{1}t)| \geq \dfrac{1}2,$$ but I am not sure if this is useful.

• Well I suppose it is definitely not continuously differentiable at $0$ at least – Maximilian Janisch Mar 22 '20 at 22:27
• In all honesty, I wasn't completely sure whether it was differentiable at 0, though I kinda felt it wasn't. – user747911 Mar 22 '20 at 23:00
• I think $H'(0) = \frac{2}{\pi}$. – GEdgar Mar 22 '20 at 23:10
• @GEdgar Elaborate please – Maximilian Janisch Mar 22 '20 at 23:15

I claim $$H'(0) = \frac{2}{\pi}$$.
Hints (When I say $$k \to \infty$$ I mean along the integers):

Step 1:
$$\lim_{k \to \infty}\frac{\displaystyle\int_{1/((k+1)\pi)}^{1/(k\pi)}\Bigg|\sin\frac{1}{t}\Bigg|\;dt}{\displaystyle\frac{1}{k\pi} - \frac{1}{(k+1)\pi}} =\frac{2}{\pi}$$

Step 2: $$\lim_{k\to\infty}k\pi\int_{0}^{1/(k\pi)} \left|\sin\frac{1}{t}\right|\;dt = \frac{2}{\pi}$$

Step 3: $$\lim_{x \to 0^+} \frac{1}{x}\int_0^x \left|\sin\frac{1}{t}\right|\;dt = \frac{2}{\pi}$$

Step 4: $$\lim_{x \to 0^-} \frac{1}{-x}\int_x^0 \left|\sin\frac{1}{t}\right|\;dt = \frac{2}{\pi}$$

Explanaton for Step 1. Write $$S_k := \frac{\displaystyle\int_{1/((k+1)\pi)}^{1/(k\pi)}\Bigg|\sin\frac{1}{t}\Bigg|\;dt}{\displaystyle\frac{1}{k\pi} - \frac{1}{(k+1)\pi}}$$ When $$k$$ is even, $$\sin\frac{1}{t} > 0$$ on the interval, when $$k$$ is odd, $$\sin\frac{1}{t} < 0$$ on the interval. We will do the even case; the odd case is similar. Change variables $$s = \frac{1}{t} - 2 k \pi$$ $$S_{2k} = \int_0^\pi\frac{(2k)(2k+1)\pi \sin(s+2 k \pi)}{(s+2 k \pi)^2}\;ds = \int_0^\pi\frac{(2k)(2k+1)\pi \sin(s)}{(s+2 k \pi)^2}\;ds$$ The integrand converges $$\lim_{k \to \infty} \frac{(2k)(2k+1)\pi \sin(s)}{(s+2 k \pi)^2} = \frac{\sin s}{\pi}\;\lim_{k \to \infty}\frac{1+\frac{1}{2k}}{1+\frac{s}{2k\pi}} = \frac{\sin s}{\pi}$$ and is dominated by $$\left|\frac{(2k)(2k+1)\pi \sin(s)}{(s+2 k \pi)^2}\right| = \frac{\sin s}{\pi}\;\frac{1+\frac{1}{2k}}{1+\frac{s}{2k\pi}} \le \frac{\sin s}{\pi}\;\frac{2}{1}$$ which is integrable on $$(0,\pi)$$. So by the dominated convergence theorem, $$\lim_{k \to \infty}S_{2k} = \int_0^\pi\frac{\sin s}{\pi}\;ds = \frac{2}{\pi}$$