With Dirac Comb is defined as follow:
$$III(t)=\sum_{n=-\infty}^\infty\delta(t-nT)$$
Fourier Transform from t domain to frequency domain can be obtained by:
$$F(f)=\int_{-\infty}^{\infty}f(t)\cdot e^{-i2\pi ft}dt$$
I wonder why directly apply the above equation does not work for the Dirac Comb:
$$F(III(t))=\sum_{n=-\infty}^\infty\int_{-\infty}^{\infty}\delta(t-nT)\cdot e^{i2\pi ft}dt$$
$$III(f)=\sum_{n=-\infty}^\infty e^{-i2\pi fnT}dt\cdots\cdots [1]$$
Where the correct way to obtain the FT of Dirac Comb is to first find the Fourier series, and then do the Fourier Transform for each term in the summation.

I copied the upper section from a open lecture slide, and I don't even understand how it goes from [2] to [3], not to mention [1] and [3] are totally not the same thing. Any hints guys?

They are equivalent. To see this, consider a function ##f(x) = \sum_n \delta (x-na)##. This is a periodic function with periodicity ##a##, which means it can be expanded in Fourier series. So
$$
f(x) = \sum_{n=-\infty}^{\infty} \delta (x-na) = \sum_{k=-\infty}^{\infty} c_k \hspace{0.5mm}e^{-i2\pi x/a}
$$
Find ##c_k## and you will see that [1] and [3] are equivalent.

That's the Fourier definition of delta function
$$
\delta(x-a) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-a)} \hspace{1mm} dk
$$

Oh I get it, with III(f) having a period of 1/T, the Fourier series should write:
$$\frac{1}{T}\sum_{n=-\infty}^\infty\delta(f-\frac{n}{T})=\sum_{k=-\infty}^\infty c_k\cdot e^{i2\pi kfT}\cdots\cdots[4]$$
$$c_k= T\cdot\int_{0}^{1/T} \sum_{n=-\infty}^\infty\delta(f-\frac{n}{T}) \cdot e^{-i2\pi kfT}=1$$

Putting c_k=1 back to [4] results in:
$$\sum_{k=-\infty}^\infty e^{i2\pi kfT} = \text{(1)}$$

Just to clear my concept, s this equivalent to (1) since the exponent missed out a -ve sign, but k ranged from -inf to inf so its just the same?