The Fourier transform of $\log^m|x|/x$ We know that $\mathcal{F}(1/x)(\xi) \sim \text{sgn}(\xi)$ is bounded. Now consider function
$$f_m(x) = \frac{\log^m|x|}{x}.$$
Can we show that $|\mathcal{F}(f_m)(\xi)| \lesssim \log^m |\xi|$? I feel we may use the Fourier transform of $\log|x|$ to get this estimate but I don't know how to carry out the details.
 A: In fact, the Fourier transform of (the principal-value integral of) $(\log |x|)^m/x$ is a constant multiple of $\mathrm{sgn}(x)\cdot (\log |x|)^m$, plus lower order terms of the form $\mathrm{sgn}(x)\cdot (\log |x|)^k$ with $0\le k<m$.
To see this, consider the family $\mathrm{sgn}(x)/|x|^s$ of tempered distributions, with $x\in\mathbb C$, initially for $\Re(s)<1$, and then by meromorphic continuation. These are odd (in the sense of parity), and homogeneous of degree $s$ (with suitable normalization of "degree"). By general properties of Fourier transforms, the Fourier transform is a constant multiple of $\mathrm{sgn}(x)/|x|^{1-s}$. Yes, for odd integer $s<0$ the function is the polynomial $x^{-s}$, and its Fourier transform is a constant multiple of an odd-order derivative of Dirac delta, which appears as a residue of the meromorphic continuation.
Let $F$ denote Fourier transform. Then
$$
F((\log |x|)^m\cdot \mathrm{sgn}(x)/|x|^{s+1})
\;=\; (-1)^m F(({\partial\over \partial s})^m \mathrm{sgn}(x)/|x|^{s+1})
\;=\; (-1)^m ({\partial\over \partial s})^m F(\mathrm{sgn}(x)/|x|^{s+1})
$$
$$
(-1)^m\cdot ({\partial\over \partial s})^m c_{s+1}\cdot (\mathrm{sgn}(x)\cdot |x|^s)
\;=\; (-1)^m\cdot \Big(c_{s+1}\cdot(\log |x|)^m+\ldots\Big)\cdot \mathrm{sgn}(x)\cdot |x|^s
$$
where $c_{s+1}$ is the constant arising in the Fourier transform. It can be determined by integrating against $xe^{-\pi x^2}$, which is multiplied by $-i$ under Fourier transform.
Evaluating at $s=0$ give
$$
F((\log |x|)^m/x) \;=\; \Big((-1)^m\cdot c_1\cdot (\log |x|)^m+\ldots\Big)\cdot \mathrm(sgn)(x)
$$
EDIT: To treat the Heaviside (=step) function $H$, multiplied by integer powers of $\log|x|$, it is perhaps clearest to break it into odd and even parts, since the family $1/|x|^s$ behaves differently than the family $\mathrm{sgn}x/|x|^s$. Certainly $2\cdot H=1+\mathrm{sgn}$. Namely, the family $\mathrm{sgn}/|x|^s$ has its first pole at $s=-2$ with residue (a constant multiple of) the derivative of Dirac $\delta$, while it is holomorphic at $s=1$ and gives (a constant multiple of) the principal-value integral against $1/x$. In contrast, the family $1/|x|^s$ has its first pole at $s=1$, and the residue is a constant multiple of Dirac $\delta$. Fourier transform does send $1/|x|^s$ to (a multiple of) $1/|x|^{1-s}$, as Fourier transform sends $\mathrm{sgn}x/|x|^s$ to a multiple of $\mathrm{sgm}x/|x|^{1-s}$. Differentiating with respect to the parameter $s$ achieves the same outcomes (and is completely justifiable via some standard devices using Gelfand-Pettis integrals and Schwartz-Grothendieck ideas about holomorphic vector-valued functions...)
