= Solution
Write $K_0(z)=-F(z)\log z+G(z)$ locally, with analytic $F,G$ and $F(0)=1$. The coefficient of $\log z$ in the <Modified Bessel differential equation> requires $z^2F''+zF'-z^2F=0$. If $F=\sum a_mz^m$, its recurrence is $m^2a_m=a_{m-2}$, with $a_1=0$. Hence
$$
F(z)=1+\frac{z^2}{4}+O(z^4),\qquad
K_0(|x|)=-\log|x|-\frac{x^2}{4}\log|x|+G(|x|)+O(x^4\log|x|).
$$
The analytic part is even: an odd-power term in $G$ would produce an uncancelled odd-power term under the same differential operator, and the recurrence removes all such terms. Insert a smooth even cutoff equal to one near zero. The smooth part and the exponentially decaying tails give no algebraic large-<wavenumber> terms after repeated <integration by parts>. Therefore the <logarithmic singularities> determine the requested terms. For $k\ne0$,
$$
\mathcal F(\log|x|)=-\frac\pi{|k|},\qquad
\mathcal F(x^2\log|x|)=-\frac{d^2}{dk^2}\left(-\frac\pi{|k|}\right)=\frac{2\pi}{|k|^3}.
$$
Thus the <Fourier transform of the modified Bessel function K0> has expansion
$$
\boxed{\widehat{K_0(|x|)}(k)=\frac\pi{|k|}-\frac\pi{2|k|^3}+O(|k|^{-5})}.
$$
An independent check follows from the <integral representation of the modified Bessel function of the second kind>. Integrating $e^{-|x|\cosh t}$ first and then setting $u=\sinh t$ gives $2\int_0^\infty du/(u^2+1+k^2)=\pi/\sqrt{1+k^2}$, with exactly these first two terms.
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