While determining Fourier coefficients we have this equation $$\int^{T}_{0} x(t) e^{-jn\omega_0t} dt = \sum^{+\infty}_{k\ =\ -\infty} a_k [\int^{T}_{0} e^{j(k-n)\omega_0t}dt]$$
I want to ask that how can we get final equation of $a_n$ as: $$a_n= 1/T \int^{T}_{0} x(t)e^{-jn\omega_0t}dt$$ from the above equation. please any one help to explain this
$$\omega_0=\frac{2\pi}{T}$$
Actually evaluate the integral on the right.
$$\int^{T}_{0} e^{j(k-n)\omega_0t}dt=\frac{T\big(e^{j(k-n)2\pi}-1\big)}{j (k-n)2\pi}=T\delta_{kn}$$
Notice that for any integer $(k-n)$ numerator is 0. So the only way we can get a non zero result if we have a $\frac{0}{0}$. The limit in that case is $T$ if $n=k$, $0$ if $k\ne n$. The special name given to this is kronecker delte $T \, \delta_{kn}$
https://en.wikipedia.org/wiki/Kronecker_delta
This means. What kronecker delta does here is the discrete equivalent of what (dirac) delta convolution does in integrals.
$$ \sum^{+\infty}_{k\ =\ -\infty} a_k [\int^{T}_{0} e^{j(k-n)\omega_0t}dt]=\sum^{+\infty}_{k\ =\ -\infty} a_kT\delta_{kn}=T a_n$$