$I\subseteq\mathbb R$ is an interval and $f,g:I\to\mathbb R$

$f$ has antiderivatives (there exists $F:I\to\mathbb R$ with $F'=f$)

$g$ is continuous on $I$

Prove that the product $f\cdot g$ has antiderivatives

This is true when $f$ is continuous or bounded, but I could neither prove it in the general case, nor find a counterexample

  • $\begingroup$ I am curious now. Have you tried guessing the antiderivative using the formula for integration by parts and checking to see if that works/what could cause it to fail? $\endgroup$ – sigma-finite Sep 22 '17 at 0:22
  • $\begingroup$ That trick only works if $g$ is differentiable, which is not necessarily the case here $\endgroup$ – Momo Sep 22 '17 at 0:55

This is not true. For example, take $I=[0,1]$ and \begin{align} f(x)=\begin{cases}\frac{1}{\sqrt{x}}\sin\frac{1}{x} & \text{for } x\in(0,1], \\ 0 & \text{for } x=0,\end{cases}\qquad g(x)=\begin{cases}\sqrt{x}\sin\frac{1}{x} & \text{for } x\in(0,1], \\ 0 & \text{for } x=0.\end{cases} \end{align} Then $f$ has an antiderivative, namely \begin{align*} F(x)=x\sqrt{x}\cos\frac{1}{x}-\frac{3}{2}\int_0^x\sqrt{t}\cos\frac{1}{t}dt, \end{align*} and $g$ is continuous, but \begin{align*} f(x)g(x)=\begin{cases}\sin^2\frac{1}{x} & \text{for } x\in(0,1], \\ 0 & \text{for } x=0,\end{cases} \end{align*} does not have an antiderivative, because it has the "wrong value" at $x=0$.


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