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The integral representation of the modified Bessel function of the second kind gives $K_0(x)={\int }_0^{\infty }{e}^{-x\cosh t}dt$ for $x>0$. Taking the Fourier transform of the even continuation and setting $u=\sinh t$ gives $2{\int }_0^{\infty }(1+k^2+u^2{)}^{-1}du=\pi /\sqrt{1+k^2}$. The large-wavenumber terms $\pi /|k|-\pi /(2|k{|}^3)$ are also obtained from the local logarithmic singularity $-\log |x|-(x^2/4)\log |x|$ using the Fourier transform of the absolute logarithm.

The barred integral in the PDF denotes a Cauchy principal value. With $p(x)=\mathrm{PV}(1/x)$ its left side is the convolution $p\ast f$, whose Fourier transform is $-i\pi \mathrm{sgn}(k)\hat{f}(k)$. The right side is $-1$ on $(-1,1)$ and zero outside; either direct integration or the Fourier transform shift rule givesThus, away from the irrelevant single point $k=0$ in the $L^2$ formulation,By the Fourier transform of the absolute logarithm, the difference $\log |x-1|-\log |x+1|$ has transform $2i\pi \sin k/|k|$; its two delta terms cancel. ThereforeIts endpoint logarithms are locally square integrable, and $f(x)=2/({\pi }^{2}x)+O(x^{-3})$ at infinity, so it belongs to L2 space. The Hilbert transform multiplier is invertible almost everywhere on this space, proving uniqueness there. Without a decay or function-space condition, one may also add a constant under the symmetric principal-value convention. If the reversed step function were used to retain the printed sign in part (i), this answer would acquire the opposite sign.

The locally integrable function $\log |x|$ defines a tempered distribution. Its distributional derivative is $\mathrm{PV}(1/x)$: perform integration by parts outside $(-a,a)$, and observe that the boundary contribution $\log a[\phi (a)-\phi (-a)]$ tends to zero. By the Fourier transform of a derivative and the preceding result,Division by $|k|$ needs a finite-part inverse-absolute-value distribution, since $1/|k|$ is not locally integrable at zero. One possible normalization isIt satisfies $k\mathrm{Pf}(1/|k|)=\mathrm{sgn}k$. The general solution of $kT=0$ is a multiple of the Dirac delta distribution: a test function vanishing at zero can be written $k\psi (k)$, so $T$ depends only on its value at zero. Consequently the Fourier transform of the absolute logarithm isChanging the finite-part normalization changes $C$. The printed reciprocal expression is meaningful with this interpretation, not as an ordinary function at $k=0$.