Let $a \in \mathbb{R}$ and $p,q$ be natural numbers with $p \geq q+2.$ For which values of $a$ is the function $$f(x) = \begin{cases} \frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}, &x \neq 0 \\ a, &x=0 \end{cases} $$ primitivable?

I noticed that $\frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}$ doesn't have an elementary antiderivative, so I tried to obtain it from the derivative of some function and work this from there. But those powers in the denominator are really getting in the way of any attempt.

I also thought of using the formula $\sin a \sin b = \frac{1}{2}(\cos(a-b)-\cos(a+b))$ and so I split the function like this: $$f(x)= \frac{1}{2}\begin{cases} \frac{1}{x}\cos(\frac{1}{x^p}-\frac{1}{x^q}), &x \neq 0 \\ a, &x=0 \end{cases} - \frac{1}{2}\begin{cases} \frac{1}{x}\cos(\frac{1}{x^p}+\frac{1}{x^q}), &x \neq 0 \\ a, &x=0 \end{cases} $$ and this definitely looks more promising, but I don't know how to proceed.

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  • Questions: 1."primitive" means "primitive on all of $\mathbb R"?$ 2. By natural number you mean an element of $\{1,2,\dots \}?$ – zhw. 7 hours ago

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