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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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  • $\begingroup$ Questions: 1."primitive" means "primitive on all of $\mathbb R"?$ 2. By natural number you mean an element of $\{1,2,\dots \}?$ $\endgroup$ – zhw. Nov 13 '18 at 21:29
  • $\begingroup$ Yes to both questions. $\endgroup$ – AndrewC Nov 14 '18 at 9:18
  • $\begingroup$ What is a primitivable function? $\endgroup$ – Mostafa Ayaz Nov 15 '18 at 18:38
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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 $$ F(x):=F(-1)+\int_{-1}^{x}f(t)\,dt. $$

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    $\begingroup$ 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 $a=0$. $\endgroup$ – N. S. Nov 16 '18 at 3:25

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