Finding the output of a system where the input is a sum of complex exponentials So, I have to find $H\{ x(t)\})$ (which is an LTI system), where $$x(t) = \sum_{k=0}^{\infty} a_ke^{ \ jw_kt}$$ and where the impulse response of the system is given by: $$h(t) = \frac{\delta(t+\tau)-\delta(t)}{\tau}$$

Here's is my attempt. I know that $$H\{x(t)\} = \int_{-\infty}^{+\infty}x(t-u)h(u)\ du \\ =\frac{1}{\tau}\left(\int_{-\infty}^{+\infty}x(t-u)\delta(u+\tau)\ du\ \  - \ \ \int_{-\infty}^{+\infty}x(t-u)\delta(u)\ du\right)$$
but I don't know where to go from there. If you could help, that would be really nice, thank you.
Okay... I managed to do it on my own. Here's the solution if you're interested. 

We know that $$H\{x(t)\} = \sum_{k=0}^{\infty}a_k \  H\{e^{jw_kt}\}$$ since $H$ is a linear system. 

We also know that $H\{e^{jw_kt}\}  = F(w_k) \ e^{jw_kt}$, where $F(w_k)$ is the frequency response of the system. 

By definition, we have that $$F(w_k) = \int_{-\infty}^{+\infty}h(t) \ e^{-jw_kt} \ dt \\ = \frac{1}{\tau} \left(\int_{-\infty}^{+\infty}\delta(t+\tau)e^{-jw_kt} dt \ - \ \int_{-\infty}^{+\infty}\delta(t) e^{-jw_kt} dt\right)\\ =\frac{1}{\tau} \left( \int_{-\infty}^{+\infty} \delta(u) e^{\ jw_k(\tau-u)} \ du \ - \ 
\int_{-\infty}^{+\infty} \delta(t)e^{\ jw_k(0-t)}\ dt \right) \\ = \frac{e^{ \ jw_k\tau} - e^0}{\tau}$$

Therefore we can simply write $$H\{x(t)\} = \sum_{k=0}^{\infty} \frac{e^{\ jw_k\tau}-1}{\tau} \ a_ke^{\ jw_kt}$$