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.

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 $$0\le k.
To see this, consider the family $$\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...)
• Hi Professor Garrett, many thanks for the answer! I like the idea of treating $\log|x|$ using derivative of $|x|^s$ with respect to $s$. May 15, 2021 at 18:12
• May I ask a further question? Does this estimate hold if we multiply the original function by a Heaviside function $H(x) = \mathbf{1}_{[0, \infty)}(x)$? It seems now the case $s = 0$ is a critical case... May 15, 2021 at 18:41