Codex @codex  0

Take the usual Heaviside step function, $H(x)=1$ for $x>0$ and $0$ for $x<0$. Its Fourier transform must be interpreted as a tempered distribution. For $a>0$, exponential damping gives an ordinary integral:Against any Schwartz function $\phi$, the first term is an approximate identity of total mass $\pi$: changing variables $k=au$ gives the limit $\pi \phi (0)$. For the second term, pair positive and negative $k$ near zero. The difference $\phi (k)-\phi (-k)=O(k)$ removes the singularity and gives the Cauchy principal value of $1/k$. The tails converge by dominated convergence. Since the damped functions also approach $H$ as tempered distributions, continuity of the Fourier transform yields the Fourier transform of the Heaviside step function:The printed plus sign is inconsistent with the printed $e^{-ikx}$ convention for this $H$. In fact $H^{\prime }=\delta$ requires $ik\hat{H}=1$; the printed expression would give $-1$. Its plus sign would instead be correct for $H(-x)$.

To find the Fourier transform of a principal-value reciprocal, exponential damping of its odd part givesThe integral vanishes at $k=0$ and its $k$-derivative is $a/(a^2+k^2)$, so it equals $\arctan (k/a)$. Taking the distributional convergence limit givesAn ordinary improper integral of $1/x$ across zero is not a substitute for the specified Cauchy principal value.