derivative is absolutely integrable

Question

I'm trying to prove that if $f$ is an infinitely differentiable function on the real line that satisfies $\lim_{|x|\rightarrow \infty}f(x)P(x)=0$ for a any polynomial $P(x)$, then it follows that:

$$\int_{-\infty}^{\infty}|f'(x)|dx<\infty$$.

I have managed to prove that $\int_{-\infty}^{\infty}|f(x)|dx<\infty$ by the following argument:

Since $\lim_{|x|\rightarrow \infty}f(x)x^{2}=0$ there exists a sufficiently large $R$ such that if $|x|>R$ then $|f(x)|<\frac{1}{x^{2}}$ and therefore:

$$\int_{-\infty}^{\infty}|f(x)|dx<\int_{-R}^{R}|f(x)|dx+\int_{|x|>R}^{.} \frac{1}{x^2}dx<\infty$$

I'm not sure how my above method can be altered in order to prove the same for the derivative.

Thank you in advance

Answer 1

Hint

Probably you want to consider a function that goes to zero exponentially, but oscillates more and more rapidly. Here is an example

$f(x) = e^{-x} \sin(e^x)$

But to prove that $\int_{-\infty}^{+\infty} |f'(x)|\,dx = +\infty$, you would maybe use a variant that is like this but has an easier antiderivative. And does the same thing going to $-\infty$.

Answer 2

Since $\lim_{|x|\rightarrow \infty}f(x)P(x)=0$ for all polynomials $P$, we have $\lim_{|x|\rightarrow \infty}(fP)'(x)=0$ (if confused as to why this holds, think about it geometrically) and thus $$ \lim_{|x|\rightarrow \infty}f'(x)P(x) =\lim_{|x|\rightarrow \infty}\left[(fP)'(x)-f(x)P'(x)\right] =\lim_{|x|\rightarrow \infty}(fP)'(x)-\lim_{|x|\rightarrow \infty}f(x)P'(x) =0 $$ for all polynomials $P$. Now apply your argument.