Derivative at $0$ of $\int_0^x \sin \frac{1}{t} dt$

Question

Let
$f(x)=\int_0^x \sin \frac{1}{t} dt \textrm{ for } x \in \mathbb R$. Is $f$ differentiable at $0$ ?

Answer 1

Let us consider the difference quotient

$$Q(x) := \frac{f(x) - f(0)}{x} = \frac1x \int_0^x \sin \frac1t\,dt$$

for small positive $x$ and apply a few substitutions:

$$\begin{align} u = \frac1t &\leadsto Q(x) = \frac1x\int_{1/x}^\infty \frac{\sin u}{u^2}\,du\\ A = \frac1x &\leadsto Q(1/A)= A \int_A^\infty \frac{\sin u}{u^2}\,du\\ u = Ay &\leadsto Q(1/A) = A\int_1^\infty \frac{\sin (Ay)}{(Ay)^2}\, d(Ay) = \int_1^\infty \frac{\sin (Ay)}{y^2}\,dy. \end{align}$$

Now, the Riemann-Lebesgue lemma says⁽¹⁾

$$\lim_{A\to\infty} Q(1/A) = 0,$$

in other words, $f$ is differentiable at $0$, with $f'(0)= 0$.

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⁽¹⁾ Riemann-Lebesgue is a bigger gun than is needed here, but it's an important thing to know, so its mention has a reason.

Here, we can do with partial integration:

$$\begin{align} \int_1^\infty \frac{\sin (Ay)}{y^2}\,dy &= \left[-\frac{\cos (Ay)}{Ay^2}\right]_1^\infty - \frac{2}{A}\int_1^\infty \frac{\cos (Ay)}{y^3}\, dy\\ &= \frac{1}{A} - \frac{2}{A} \int_1^\infty \frac{\cos (Ay)}{y^3}\, dy \end{align}$$

and we can estimate

$$\left\lvert \int_1^\infty \frac{2\cos (Ay)}{y^3}\,dy\right\rvert \leqslant \int_1^\infty \frac{2}{y^3}\,dy = 1$$

to obtain

$$\lvert Q(1/A)\rvert \leqslant \frac{2}{A}.$$