Fourier transform of shifted periodic function

Question

Assuming $x(t)$ is a periodic function of period $T$ and having the Fourier transform $X(\omega)$, it is required to calculate the Fourier transform of the signal $x(t)+x(t-T)$. Since x(t-T) is equal to x(t) the Fourier transform should simply be $2X(\omega)$ but if we use the time-shifting property of the Fourier transform the answer should also be $X(\omega)+e^{-j\omega T} X(\omega)$. But how come I am getting two different answers.

Actually, I am confused about the concepts involving the Fourier transform of periodic signals and while practicing I came up with this weird thing and it's getting more confusing, and chances are it may not make sense to some people but I don't know why two fully applicable things giving different conclusions.

Answer 1

It's important to realize that a $T$-periodic function has a discrete frequency spectrum with contributions at integer multiples of $\omega_0=2\pi/T$. Consequently, the spectrum $X(\omega)$ has the form

$$X(\omega)=\sum_kc_k\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{1}$$

with constants $c_k$, which are just scaled versions of the Fourier coefficients of $x(t)$.

When multiplied by $e^{-j\omega T}$ one obtains contributions

$$e^{-j\omega T}\delta\left(\omega-\frac{2\pi k}{T}\right)= e^{-j2\pi k}\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{2}$$

because for any $f(\omega)$ that is continuous at $\omega_k$ we have $f(\omega)\delta(\omega-\omega_k)=f(\omega_k)\delta(\omega-\omega_k)$. Furthermore, since $e^{-j2\pi k}=1$ the result follows, i.e.,

$$e^{-j\omega T}X(\omega)=X(\omega)\tag{3}$$