= Solution
Interpret the <Fourier transform> as a <distributional Fourier transform>. Exponentially regularize the <Heaviside step function>:
$$
\widehat{e^{-ax}H(x)}(k)=\int_0^\infty e^{-(a+ik)x}dx=\frac1{a+ik}=\frac a{a^2+k^2}-i\frac k{a^2+k^2},\qquad a>0.
$$
As $a\downarrow0$, the first term tends to $\pi\delta(k)$ in the <tempered distribution> sense, since its integral is $\pi$ and it concentrates at zero. The second tends to $-i\operatorname{PV}(1/k)$. Therefore
$$
\boxed{\widehat H(k)=\pi\delta(k)+\operatorname{PV}\frac1{ik}.}
$$
The principal-value convention is essential: the printed $1/(ik)$ is not an ordinary integrable function at zero. This normalization also satisfies $ik\widehat H=1$, the <Fourier transform of a derivative> version of $H'=\delta$.
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