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.
Best Answer
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, $$ \lim_{k \to \infty}S_{2k} = \int_0^\pi\frac{\sin s}{\pi}\;ds = \frac{2}{\pi} $$