Calculate the Fourier Transform of Unit Step Function $u(t)$ Directly

Question

My deduction stops here: $$X(j\omega)=\int_{0}^{+\infty}e^{-j\omega t}\mathrm{d}t =\left.\frac{1}{j\omega}e^{-j\omega t}\right|_{0}^{+\infty}=\frac{1}{j\omega}\left(1-\lim_{N\rightarrow+\infty}e^{-j\omega N}\right)=\frac{1}{j\omega}-\lim_{N\rightarrow+\infty}\frac{\cos(\omega N)-j\sin{\omega N}}{j\omega}=\frac{1}{j\omega}-\lim_{N\rightarrow+\infty}\frac{\cos(\omega N)}{j\omega}+\lim_{N\rightarrow+\infty}\frac{\sin(\omega N)}{\omega}$$ Note that$$\lim_{N\rightarrow+\infty}\frac{\sin(\omega N)}{\omega}=\pi\delta(\omega)$$ (This property is directly given in DSP course without proof. If possible, please also give its proof process, thanks a lot)

But how can I calculate $\lim_{N\rightarrow+\infty}\frac{\cos(\omega N)}{j\omega}$ ? Or as the answer shows, how can I prove that $\lim_{N\rightarrow+\infty}\frac{\cos(\omega N)}{j\omega}=0$ ?

By the way, I have read and understood the solution using $sgn(t)$, I'm just wondering how to continue the derivation with the above method.

Answer 1

The integral $\int_{0}^\infty e^{-i\omega t}\,dt$ fails to exist as a Riemann or Lebesgue integral. So, the Fourier transform of the unit step function must be considered in the context of a distribution.

Here, we assume that $\phi\in \mathbb{S}$. Then, we have

$$\begin{align} \langle \mathscr{F}\{u\}, \phi\rangle&=\langle u, \mathscr{F}\{\phi\}\rangle\\\\ &=\int_0^\infty \int_{-\infty}^\infty \phi(t)e^{-i\omega t}\,dt\,d\omega\\\\ &=\lim_{L\to \infty}\int_0^L \int_{-\infty}^\infty \phi(t)e^{-i\omega t}\,dt\,d\omega\\\\ &=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(t)\int_0^L e^{-i\omega t}\,d\omega\,dt\\\\ &=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(t)\frac{1-e^{-i Lt}}{it}\,dt\\\\ &=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(t)\frac{1-\cos(Lt)}{it}\,dt+\lim_{L\to \infty}\int_{-\infty}^\infty \phi(t)\frac{\sin(Lt)}{t}\,dt\tag1 \end{align}$$

Now, we cannot simply interchange the order of the limit and the integral. Instead, we proceed as follows.

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For the first integral on the right-hand side of $(1)$, we can write for any $\delta>0$

$$\begin{align} \int_{-\infty}^\infty \phi(t)\frac{1-\cos(Lt)}{it}\,dt&=\int_{|t|\le \delta} \phi(t)\frac{1-\cos(Lt)}{it}\,dt+\int_{|t|\ge \delta} \phi(t)\frac{1-\cos(Lt)}{it}\,dt\\\\ &=\underbrace{\int_{|t|\le \delta} (\phi(t)-\phi(0))\frac{1-\cos(Lt)}{it}\,dt}_{=O(\delta)}\\\\ &+\phi(0)\underbrace{\int_{|t|\le \delta} \frac{1-\cos(Lt)}{it}\,dt}_{=0\,\,\text{since the integrand is odd}}\\\\ &+\int_{|t|\ge \delta} \frac{\phi(t)}{it}\,dt\\\\ &-\underbrace{\int_{|t|\ge \delta} \phi(t)\frac{\cos(Lt)}{it}\,dt}_{=O(1/L)\,\,\text{from integrating by parts}}\tag2 \end{align}$$

Hence, letting $L\to \infty$ first, and then $\delta\to 0^+$ we find that

$$\lim_{L\to \infty}\int_{-\infty}^\infty \phi(t)\frac{1-\cos(Lt)}{it}\,dt=\text{PV}\int_{-\infty}^\infty \frac{\phi(t)}{it}\,dt\tag3$$

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For the second integral on the right-hand side of $(1)$, we have by integrating by parts with $u=\phi(t)$ and $v=\int_{-\infty}^{Lt} \frac{\sin(t')}{t'}\,dt'$ and applying the Dominated Convergence Theorem

$$\begin{align} \lim_{L\to\infty}\int_{-\infty}^\infty \phi(t)\frac{\sin(Lt)}{t}\,dt&=-\lim_{L\to\infty}\int_{-\infty}^\infty \phi'(t)\int_{-\infty}^{Lt} \frac{\sin(t')}{t'}\,dt'\,dt\\\\ &= \int_{-\infty}^\infty \phi'(t)\lim_{L\to\infty}\int_{-\infty}^{Lt} \frac{\sin(t')}{t'}\,dt'\,dt\\\\ &= \int_{-\infty}^\infty \phi'(t) \pi u(t)\,dt\\\\ &=\pi \phi(0)\tag4 \end{align}$$

Putting together $(3)$ and $(4)$ in $(1)$ we find that in distribution

$$\mathscr{F}\{u\}(\omega)=\pi\delta(\omega)+\text{PV}\left(\frac1{i\omega}\right)$$