Full Proof of Finding Fourier Coefficients For an Impulse Train Firstly please keep this in mind(complex fourier series):
$x(t)= \sum\limits_{k=-\infty}^{\infty} c_k \times e^{J\frac{2\pi kt}{T}} $ <==>
$c_k= \frac{1}{T}\times\int\limits_{t=-\infty}^{\infty} x(t) \times e^{-J\frac{2\pi kt}{T}} dt$


Let $x(t)=\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)$ which is an impulse train.
Plug this on the formula which is on right:
$c_k= \frac{1}{T}\times\int\limits_{t=-\infty}^{\infty} [\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)] \times e^{-J\frac{2\pi kt}{T}}dt $

As n and t are independent variables, we can interchange order of the integral and sum:
$c_k=  \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}[\int\limits_{t=-\infty}^{\infty}\delta(t-nT) \times e^{-J\frac{2\pi kt}{T}}]dt $

getting rid of the integral using the sifting property:
$c_k=  \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}[ e^{-J\frac{2\pi knT}{T}}]dt =  \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}[ e^{-J2\pi kn}]dt $

factoring out, using property $a^{(b\times c )}=(a^b)^c$:
$c_k = \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}{(e^{-J2\pi n})}^k = \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty} 1^k=\infty$
We have $c_k = \infty$ but we know that $c_k = \frac{1}{T}$ from books.
So, where is my mistake and how can I correct it?
I was studying Signals&Systems but I'm stuck, need your help.
I appreciate your suggestions.
Thank you so much.
 A: Thanks for your help,
The first relation was wrong, right one is:
$x(t)= \sum\limits_{k=-\infty}^{\infty} c_k \times e^{J\frac{2\pi kt}{T}}$<==>
$c_k= \frac{1}{T}\times\int\limits_{t=-T/2}^{T/2} x(t) \times e^{-J\frac{2\pi kt}{T}} dt $

let $x(t)=\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)$.
Plug it on right formula:
$c_k= \frac{1}{T}\times\int\limits_{t=-T/2}^{T/2} [\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)] \times e^{-J\frac{2\pi kt}{T}}dt$

change order of the integral and sum:
$c_k= \frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}[\int\limits_{t=-T/2}^{T/2} \delta(t-nT) \times e^{-J\frac{2\pi kt}{T}}dt]$
call the new integral as $I$:
$I=\int\limits_{t=-T/2}^{T/2} \delta(t-nT) \times e^{-J\frac{2\pi kt}{T}}dt$
$I$ is a partial function of integers $n$ and $k$, we can easily decompose it using sifting property.
So, the integral has only one nonzero value at given range as we see:
$I=\begin{cases} 
      1 & n=0 \\
      0 & n\neq0  \\
   \end{cases}$
(As wee saw, $I$ is independent of $k$)

recall: $c_k=\frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}[\int\limits_{t=-T/2}^{T/2} \delta(t-nT) \times e^{-J\frac{2\pi kt}{T}}dt]$
replace integral with I:
$c_k=\frac{1}{T}\times\sum\limits_{n=-\infty}^{\infty}I$
as $I$ is nonzero at only $n=0$:
$c_k=\frac{1}{T}$
