# What is incorrect in my way for getting Fourier transform of step function?

Today I tried to get Fourier transform of step function ($$u(t)$$). But I got a result which seems is not correct. I want to know what is incorrect in my work?

With attention to this relation: \begin{align} \mathrm{u(t)=\frac{1}{2}[1+sgn(t)]} \end{align} we know that: \begin{align} \mathfrak{F}(u(t))=\left (\pi\delta(\omega) + \frac{1}{j\omega}\right ) \end{align}

Today I tried to experience another way. Suppose:

\begin{align} g(t) = \begin{cases} e^{-at} & \text{for } t > 0\\ 0 & \text{otherwise } \end{cases} \end{align}

So:

\begin{align*} \mathfrak{F}(g(t)) = &\int_{0}^{\infty} e^{-at} e^{-i\omega t}\, dt\\ &=\int_{0}^{\infty}e^{-(a+i\omega) t}\, dt\\ &=\big[ \frac{e^{-(a+i\omega) t}}{-(a+i\omega)} \big]_{0}^{\infty}\\ &=\frac{1}{a+i\omega} \end{align*}

It is obvious that $$\lim\limits_{a\to 0^+}g(t)= u(t)$$. Thus:

\begin{align} \mathfrak{F}(u(t)) &= \mathfrak{F}(\lim\limits_{a\to 0^+}g(t))\\ &=\lim\limits_{a\to 0^+}(\mathfrak{F}(g(t))\\ &=\lim\limits_{a\to 0^+}(\frac{1}{a+i\omega})\\ &=\frac{1}{i\omega} \end{align} What is incorrect in my work?

• The integral of the limit is not always the limit of the integral. Apr 12, 2023 at 22:15
• @ThomasAndrews Well, but when it is correct? I saw that in getting Fourier transform of sgn function. Apr 12, 2023 at 22:18
• You need to prove it is correct to switch the limit and integral. Here, of course, the only place where it is incorrect is when $\omega=0.$ Apr 12, 2023 at 22:32
• But the real problem is that $1/(a+wi)$ does not converge to anything when $\omega=0.$ Apr 12, 2023 at 22:34
• To add to the comment posted by @Gonçalo , the Fourier transform is a distribution on the Schwartz space $\mathbb{S}$ of functions. The limit $\lim_{a\to 0^+}\left(\frac1{a+i\omega}\right)$ is a distributional limit, such that for any $\phi\in \mathbb{S}$, we have $$\lim_{a\to 0^+} \int_{-\infty}^\infty \frac{\phi(\omega)}{a+i\omega}\,d\omega=\pi \phi(0)+\text{PV}\int_{-\infty}^\infty \frac{\phi(\omega)}{i\omega}\,d\omega$$ Apr 13, 2023 at 13:23