Prove that the Fourier transform of a function agreeing with $1/\log(|x|)$ for $|x| \geq 2$ is integrable. 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?
 A: 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.
