Note:$$j=i=\sqrt{-1}$$
The Fourier Transform series coefficient formula: $$X(jw)=\int^{\infty}_{-\infty}x(t)e^{-j\omega t}dt$$
We choose $$x(t)=e^{j\omega_0 t}$$
$$X(jw)=\int^{\infty}_{-\infty}e^{j{w_0-w}t}=\frac{e^{j(w_0-w)\infty}-e^{j(w_0-w)-\infty}}{j(\omega_0-\omega)t}=\frac{\infty-0}{j(\omega_0-\omega)t}$$
I know the right answer is $2\pi\delta(\omega-\omega_0)$. But how do you see it?