# For which values of $a$ is $f$ primitivable?

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