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I was trying to derive the fourier transform of the step function explicitly, and i came up with doing the same reasonements we can see in the askers' attempts in their respective questions:

In both cases, the askers are struggling to find a proof of the basic intuition: $$\left[\lim_{x\to +\infty} \frac{\cos(kx)}{ik}=0\right]$$ At this point, my questions are:

  1. It seems that the former intuition came from the statement:
    "fourier transform of step function" $= \frac{1}{ik}+\pi \delta(k)$
    Is this true? I read some sources that indicates $= PV\left(\frac{1}{ik}\right)+\pi \delta(k)$
    whatever "$PV\left(\frac{1}{ik}\right)$" means..
  2. The result: $\left[\lim_{x\to +\infty} \frac{\sin(kx)}{k}=\pi \delta(k) \right]$ is straightforward, since it is the very definition of "$\delta(k)$". Indeed, i can show that $$\int_{-\infty}^{+\infty} dx \frac{\sin(nx)}{\pi x} = 1, \quad \forall n \in \mathbb{N}$$ $$\lim_{n \to \infty} \int_{-\infty}^{+\infty} dx \frac{\sin(nx)}{\pi x} f(x) = f(0)$$ In this sense, we have demonstarte that $\left[\lim_{x\to +\infty} \frac{\sin(kx)}{k}\right]$ has the properties of a delta. But what can we say about $\left[\lim_{x\to +\infty} \frac{\cos(kx)}{ik}\right]$? Has it the "same properties" of zero?
TheRvela
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    The step function is not in $L^1$ or $L^2$, so one has to use the Fourier transform in the sense of distributions. The first link you indicate did the computation carefully, what do you not understand there? $\mathrm{pv}(1/x)$ is the principal value of $1/x$ (look any course on distribution theory) – LL 3.14 Mar 16 '23 at 09:08

2 Answers2

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Using the formulas $$ \mathcal{F}\{\operatorname{sign}(t)\} = -2i\operatorname{pv}\frac{1}{\omega} $$ and $$ \mathcal{F}\{1\} = 2\pi\delta(\omega) $$ we get $$ \mathcal{F}\{u(t)\} = \mathcal{F}\{\frac12(1+\operatorname{sign}(t))\} = \frac12\left(2\pi\delta(\omega)-2i\operatorname{pv}\frac{1}{\omega}\right) = i\operatorname{pv}\frac{1}{\omega} + \pi\delta(\omega). $$

md2perpe
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The Fourier transform of the unit step function $u(t)$ has proven to be a delicate case; a direct calculation as you intended to do turns out to be a nightmare of regularization and abuse of notation. One way to circumvent those issues relies on the fact that the derivative of the Heaviside function is precisely the Dirac delta function, i.e. $\dot{u}(t) = \delta(t)$. In consequence, the properties of the Fourier transform leads to $i\omega\,\hat{u}(\omega) = 1$. However, one needs to handle this equation carefully, because one looks for solutions in a distribution space.

The naive solution is given by $\hat{u}(\omega) = \frac{1}{i\omega}$. This expression is well defined for $\omega \neq 0$ and corresponds to the usual inverse function, while it behaves as a Dirac delta at $\omega = 0$. Indeed, $\frac{1}{2\pi i\omega}$ is a representation of the Dirac delta on the complex plane due to Cauchy's integral formula $-$ which will be used when computing the inverse Fourier transform for instance. Consequently, one has $\hat{u}(\omega) = \frac{1}{i\omega} = PV\frac{1}{i\omega} + \pi\delta(\omega)$; note that it lacks a factor $2$, because the singularity $\omega = 0$ lies on the real line. You can see this as a decomposition of $\hat{u}(\omega)$ between its regular part (because the Cauchy principal value is equivalent to ignore the singularity) and its singular part at $\omega = 0$.

It is to be noted too that the distributional equation $\omega\,\hat{u}(\omega) = -i$ can be solved in a algebraic style. In a standard setup, it corresponds to a linear equation with the unique (particular) solution $\frac{1}{i\omega}$ (which has to be understood as a principal value), because the multiplication operator $F(\omega) \mapsto \omega F(\omega)$ has a trivial kernel (and thus is invertible). In a distributional space however, its kernel is not trivial anymore; indeed, its is spanned by the Dirac delta because of the well-known relation $\omega\delta(\omega) \equiv 0$. As a consequence, one has to add the general solution of the associated homogeneous equation $\omega \,\hat{u}(\omega) = 0$ to the aforementioned particular solution, hence finally $\hat{u}(\omega) = PV\frac{1}{i\omega} + c\delta(\omega)$, with $c$ a constant. This constant will have to be determined by matching the inverse Fourier transform with the initial unit step function $u(t)$.

Finally, since the Fourier transform $\hat{f}(\omega)$ can be seen as the limit of the Laplace transform $F(s)$, more precisely $\hat{f}(\omega) = \displaystyle \lim_{\sigma\rightarrow0} F(\sigma + i\omega)$, we can also recover the Fourier transform of the unit step function by applying the Sokhotski–Plemelj theorem to its Laplace transform $U(s) = 1/s$.

Abezhiko
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