I am looking at a question which states that if $f(t) = \sum_{n = -\infty}^{\infty} {c_n e^{j n w_0 t}}$ and $g(t) = \sum_{n = -\infty}^{\infty} {d_n e^{j n w_0 t}}$ where $w_0 = \frac{2 \pi}{T}$ then show that $f(t) \star g(t) = \sum_{n = -\infty}^{\infty} {c_n d_n e^{j n w_0 t}}$ where $\star$ is convolution.
I get a different result: $f(t) \star g(t) = 2 \pi \sum_{n = -\infty}^{\infty} {c_n d_0 e^{j n w_0 t}}$
My work:
$$ f(t) \star g(t) = \int_{-\infty}^{\infty} {\sum_{n = -\infty}^{\infty} {c_n e^{j n w_0 t}} \sum_{k = -\infty}^{\infty} {d_k e^{j k w_0 (t-\tau)}}}d \tau = \sum_{n = -\infty}^{\infty} {c_n e^{j n w_0 t}} \sum_{k = -\infty}^{\infty} \left[ {d_k e^{j k w_0 t}} \int_{-\infty}^{\infty} {e^{-j k w_0 \tau}} d \tau \right] $$
Now I use the fact that $ \int_{-\infty}^{\infty} {e^{-j w \tau}} d \tau $ (note $\omega$ instead of $k \omega_0$ is the Fourier Transform of 1 and so this integral is equal to $\sqrt{2 \pi} \delta(\omega)$. This means that $ \int_{-\infty}^{\infty} {e^{-j k w_0 \tau}} d \tau $ (note $k \omega_0$ instead of $\omega$) is equal to $\sqrt{2 \pi} \delta(k \omega_0)$ which is 0 for $k \ne 0$ and $\sqrt{2 \pi}$ for $k = 0$.
Therefore,
$$ \sum_{k = -\infty}^{\infty} \left[ {d_k e^{j k w_0 t}} \int_{-\infty}^{\infty} {e^{-j k w_0 \tau}} d \tau \right] = \sum_{k = -\infty}^{\infty} {d_k e^{j k w_0 t}} 2 \pi \delta(k \omega_0) = 2 \pi d_0 $$
Thus,
$$ f(t) \star g(t) = 2 \pi \sum_{n = -\infty}^{\infty} {c_n d_0 e^{j n \omega_0 t}} $$