Why does the first term in the integral $\int_{-\infty}^\infty f'(x)e^{-i \omega x}dx$ vanish?

Question

Given $f \in L^1(\mathbb{R})\cap C^1(\mathbb{R})$, I'm trying to solve the following integral using integration by parts:

$$\int_{-\infty}^\infty f'(x)e^{-i \omega x}dx$$

I know that the final result should be the second term in the integration by parts (below), but I cannot see why the first part vanishes (note, I may be a little rusty on improper integrals).

Here's what I've done:

$$ \int_{-\infty}^\infty f'(x)e^{-i \omega x}dx = \lim_{a \to \infty} \bigg(e^{-i\omega x }f(x) \bigg|_{-a}^{a} \bigg ) +i\omega \int_{-\infty}^{\infty}e^{-i\omega x} f(x) dx$$

Why is the first term $0$ (it definitely has to be because this integral is a known identity for Fourier Transforms)? What am I doing wrong / not seeing? When I plug in the limits, nothing seems to vanish (I used Euler's identity to transition to $sin,cos$ and I'm just not seeing it).

Thanks!

Answer 1

No, $f\in L^1\cap C^\infty$ does not imply that $f$ vanishes at infinity. But

Lemma If $f\in L^1(\Bbb R)$ then there exists a sequence $(a_n)$ with $a_n\to\infty$ and $|f(a_n)|+|f(-a_n)|\to0$.

That's enough to do the integration by parts, which is enough to show that $\lim_{a\to\infty}\int_{-a}^af'(t)e^{i\omega t}$ is what it should be (for $f\in L^1\cap C^1$)...