Let $f \in C^\infty(\mathbf{R})$ be an even, smooth function such that for $|x| \geq 2$, $f(x) = 1/\text{log}|x|$. I am trying to show that $\widehat{f}$ is integrable. This must use the fact that the function is even, because odd functions agreeing with $1/\text{log}(x)$ for $x \geq 1$ do not have an integrable Fourier transform.
To try and understand this problem, we consider a decomposition in space and frequency. Define $f_j(x) = \chi_j(x) f(x)$, where $\chi_j(x) = \chi(x/2^j)$, and $\chi$ is a smooth cutoff supported on $|x| \sim 1$. We then need to understand the quantities $\| \chi_i \widehat{f_j} \|_{L^1}$ for $j \geq 0$, and $i \in \mathbf{Z}$.
Simply by calculating the derivative of $f$, and then applying Parseval's identity, I was able to show that $\| \chi_i \widehat{f_j} \|_{L^1} \lesssim 2^{-(i+j)/2} / j^2$. This gives good bounds for $i + j \geq 0$. My problem is obtaining bounds on the quantities $\| \chi_i \widehat{f_j} \|_{L^1}$ when $i + j \leq 0$. Since we haven't used the fact that $f$ is even, this must be where this property becomes useful. But I can't quite see how to do this.
Intuitively, to get good bounds we need to understand the values of $\widehat{f_j}(\xi)$ where $|\xi| \leq 1/2^j$. The fact that we have to exploit the fact that $f$ is even suggests using the identity
$$ \widehat{f_j}(\xi) = \int \frac{\chi(x/2^j)}{\log |x|} e^{2 \pi i \xi \cdot x} dx = 2 \int_0^\infty \frac{\chi(x/2^j)}{\log |x|} \cos(\xi x) dx. $$
Taking in absolute values gives $|\widehat{f_j}(\xi)| \lesssim 2^j / j$, and thus $\| \chi_i \widehat{f_j} \|_{L^1} \lesssim 2^{i+j} / j$, but this is not strong enough to sum over all $i + j \leq 0$. Nothing is really oscillating in the integral, so it does not seem like integration by parts will help us. What can we do?
Credit to Maximilian Janisch for suggesting this approach.
Based on the approach suggested in this problem, it suffices to show the function
$$ g(x) = \sum_{j = 0}^\infty \sum_{i = -\infty}^{-j-10} \chi_i \widehat{f_j} $$
is integrable. Using the fact that $f_j$ is even, we find that
$$ \widehat{f_j}(\xi) = 2 \int_2^\infty \cos(2 \pi \xi x) \frac{\chi_j(x)}{\log(x)}\; dx. $$
All quantities here are non-negative for $|\xi| \leq 1/2^{-j-10}$ since then $|\xi x| \leq 2^{-10}$. Thus we can apply Tonelli's theorem to conclude that
$$ \| g \|_{L^1} \lesssim \sum_{i = -\infty}^{0} \sum_{j = i + 10}^\infty \int_2^\infty \int \chi_i(\xi) \cos(2 \pi \xi x) \frac{\chi_j(x)}{\log(x)}\; dx\; d\xi \lesssim \int_{-2^{-10}}^{2^{-10}} \int_2^{2^{-10}/|\xi|} \frac{\cos(2 \pi \xi x)}{\log(x)}\; dx\; d\xi. $$
But this integral can also be written as
$$ \int_2^\infty \frac{1}{\log(x)} \int_{-2^{-10}/x}^{2^{-10}/x} \cos(2 \pi \xi x)\; d\xi\; dx = \int_2^\infty \frac{\sin(2 \pi \cdot 2^{-10}))}{\pi x \log(x)} $$
But this seems to show that $g$ is not integrable, since these inequalities are essentially sharp, and $1/x log(x)$ is a divergent integral.