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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 $z^2{F}^{\prime \prime }+z{F}^{\prime }-z^{2}F=0$. If $F=\sum a_m{z}^m$, its recurrence is $m^2{a}_m=a_{m-2}$, with $a_1=0$. HenceThe 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 $k\ne 0$,Thus the Fourier transform of the modified Bessel function K0 has expansionAn 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.