Let me denote Fourier transform by $\mathcal F.$
Proposition
Let $f:\mathbb R\to \mathbb C$ is twice differentiable and suppose $f,f',f''\in L^1\cap C$ and $f(x), f'(x)\to 0$ as $|x|\to\infty.$ Then, $\mathcal Ff\in L^1.$
Here is the proof written in my book.
Proof
By integrating by parts, for $\xi\neq 0$, we get \begin{align} |\mathcal F f(\xi)| =|\int_{-\infty}^\infty f(t)e^{-i\xi t}dt| \leqq A\frac{1}{|\xi|^2}, \end{align} where $A=\int_{-\infty}^\infty |f''(x)|dx.$
Thus $\mathcal F f\in L^1$. Q.E.D.
I don't know why he says $\mathcal Ff\in L^1$ from $|\mathcal Ff(\xi)|\leqq A\frac{1}{|\xi|^2}$ for $\xi\neq 0$.
$\int_{-\infty}^\infty \frac{1}{|\xi|^2}d\xi$ doesn't converge, so we cannot say $\int_{-\infty}^\infty |\mathcal Ff(\xi)|d\xi<\infty$ by comparison.
Indeed, does this proof work ? If not, can we prove by another way, or is this proposition false ?
Break up your integral into $\int_{|\xi|\leq 1}+\int_{|\xi|>1}$ for instance. The Fourier transform of $f\in L^1$ is continuous everywhere. So near the origin use continuity, and away from the origin, use the decay rate you found. In general we always use the estimate which is better suited to the region involved (a recurring theme with Fourier/Harnonic analysis).