# 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.

• Questions: 1."primitive" means "primitive on all of $\mathbb R"?$ 2. By natural number you mean an element of $\{1,2,\dots \}?$ – zhw. Nov 13 '18 at 21:29
• Yes to both questions. – AndrewC Nov 14 '18 at 9:18
• What is a primitivable function? – Mostafa Ayaz Nov 15 '18 at 18:38

Take $$a=0$$. Consider the function $$G(x):=x^{p}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}.$$ Then for $$x\neq0$$, \begin{align*} G^{\prime}(x) & =px^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}-px^{-1}% \sin\frac{1}{x^{p}}\sin\frac{1}{x^{q}}\\ & -qx^{p-q-1}\cos\frac{1}{x^{p}}\cos\frac{1}{x^{q}}\\ & =:h(x)-pf(x), \end{align*} while for $$x=0$$, $$\frac{G(x)-G(0)}{x-0}=x^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}% \rightarrow0$$ since $$p\geq2$$. Now the function $$h(x)=px^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}-qx^{p-q-1}\cos\frac {1}{x^{p}}\cos\frac{1}{x^{q}}%$$ is continuous at $$x=0$$ provided we set $$h(0):=0$$, since $$p-q-1\geq1$$ and so $$\lim_{x\rightarrow0}h(x)=0.$$ It follows that $$h$$ has an antiderivative $$H(x):=\int_{0}^{x}h(t)\,dt$$. It follows that$$-pf(x)=G^{\prime}(x)-H^{\prime}(x)$$ and so the function given for $$x\in\lbrack-1,1]$$ by$$F(x):=-\frac{1}{p}G(x)+\frac{1}{p}H(x)$$ has derivative $$F^{\prime}(x)=-\frac{1}{p}G^{\prime}(x)+\frac{1}{p}H^{\prime }(x)=f(x)$$.
For $$x\geq1$$ define $$F(x):=F(1)+\int_{1}^{x}f(t)\,dt.$$ Note that $$f$$ is integrable in $$[1,\infty)$$ since $$\sin t\sim t$$ for $$t\rightarrow0$$ and so as $$x\rightarrow\infty$$ we have $$\frac{1}{x}\sin\frac{1}{x^{p}}\sin\frac{1}{x^{q}}\sim\frac{1}{x^{1+p+q}}$$ and $$1+p+q\geq3$$. Then $$F^{\prime}(x)=f(x)$$ for all $$x\geq1$$.
Similarly, for $$x\leq-1$$ define $$F(x):=F(-1)+\int_{-1}^{x}f(t)\,dt.$$
• To complete the answer just observe that if $g(x)=\begin{cases} \frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}, &x \neq 0 \\ a, &x=0 \end{cases}$ has a primitive, then so does $$f-g=\begin{cases} 0, &x \neq 0 \\ a, &x=0 \end{cases}$$But any function having a primitive satisfies the intermediate value property, thus $a=0$. – N. S. Nov 16 '18 at 3:25