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

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

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