# Does the product $fg$ have antiderivative?

Let $$I$$ be an interval and $$g$$ be a continuous function $$g\colon I\to\mathbb{R}$$. Suppose that $$f\colon I\to\mathbb{R}$$ is a nonzero function and $$f$$ has antiderivative. Show that the product $$fg$$ has antiderivative.

I tried to use the fact that $$fg=\frac{\left(f+g\right)^{2}-f^{2}-g^{2}}{2}$$ and since $$g$$ is continuous then $$g^{2}$$ is also continuous and therefore it has antiderivative. Also $$f+g$$ has antiderivative, but I don't know if $$f^{2}$$ and $$\left(f+g\right)^{2}$$ have antiderivative. I couldn't prove that the square of these functions has antiderivative and also I don't know if this statement is false. Could you give me some suggestion?

Thanks.

• If you're trying to show that $fg$ has an elementary antiderivative, it doesn't. Counterexample: $f(x)=1,$ and $g(x)=e^{-x^2}.$ If you're trying to show that $fg$ is integrable, then please change your wording, as that's quite different. – Adrian Keister Jun 28 '19 at 18:53
• @AdrianKeister, the OP's terminology seems OK to me. See, e.g., en.wikipedia.org/wiki/Antiderivative – Barry Cipra Jun 28 '19 at 18:56
• @AdrianKeister No, I only have to show that $fg$ has antiderivative, but not that is elementary. – oioa Jun 28 '19 at 19:07
• The condition that $f$ is not zero must be relevant, compare math.stackexchange.com/q/2439669/42969. – Martin R Jun 28 '19 at 19:09
• @MartinR May be that's important. Even so I don't know how can I prove that the product has antiderivative. Is there a sufficient condition to prove that a function has antiderivative? Since there are counterexamples that show that the function could not have an elementary antiderivative. – oioa Jun 28 '19 at 19:30

Short answer: If $$F$$ is the antiderivative of $$f$$, then: $$\int g \cdot f = \int g\circ \mathbb{1}_{I} \cdot f = \int g\circ F^{-1}\circ F \cdot f = \left(\int g\circ F^{-1}\right)\circ F$$

Let $$F$$ is the antiderivative of $$f$$. Obviously $$F$$ is continuous.

By Darboux's theorem, $$f$$ has the intermediate value property.

As $$f(x)\ne 0$$, it follows that either $$f>0$$ or $$f<0$$ (otherwise by intermediate value property $$f$$ would take the value zero).

It follows that $$F$$ is strictly monotone, therefore injective.

Let $$J$$ be the image of $$F$$. By intermediate value property, $$J$$ is an interval. Moreover, $$F:I\to J$$ is continuous and injective, so it has a continuous inverse $$F^{-1}:J\to I$$

Let $$h:J\to \mathbb{R}$$ with $$h=g\circ F^{-1}$$. $$h$$ is continuous as composition of two continuous functions, thus it has an antiderivative $$H:J\to \mathbb{R}$$

Then $$H\circ F:I\to \mathbb{R}$$ is differentiable, as composition of two differentiable functions, and by the chain rule: $$(H\circ F)'=H'\circ F\cdot F'=H'\circ F\cdot f=g\circ F^{-1}\circ F\cdot f=g\circ\mathbb{1}_I\cdot f=g\cdot f$$

That's all, folks!