I need some help with simplifying the following expression:
$$\sum_{n= -\infty }^\infty \delta(t-2n) \ast e^{-2t}u(t)$$
Using the convolution integral and taking the unit step into account, I get:
$$\int_{0}^\infty\sum_{n=0 }^\infty \delta(t-2n)e^{-2{(t-\tau)}}d\tau $$
Simplifying further results in:
$$e^{-2t}\sum_{n=0 }^\infty \delta(t-2n)\int_{0}^\infty e^{2\tau}d\tau $$
The integral $\int_{0}^\infty e^{2\tau}d\tau $ does not converge, so I'm unsure if I made a mistake somewhere or if this is what the expression $\sum_{n= -\infty }^\infty \delta(t-2n) \ast e^{-2t}u(t)$ simplifies to.
Can someone verify my work or point me in the right direction?