On my midterm, I had the following question:
Calculate the Laplace transform of $$\sum_{k=0}^{+\infty}(-1)^{k}\delta (t-k)$$
I was wondering how I should calculate it. I know that the transform would be $\dfrac{e^s}{1+e^s}$, and we were given the hint that we may need to use
$$\sum_{k=0}^{+\infty} a^{k} = \dfrac{1}{1-a}$$
for $|a|<1$. My original thought was to write out the sum, which would become
$$\sum_{k=0}^{+\infty}(-1)^{k}\delta (t-k) = [\delta(t)-\delta(t-1)] + [\delta(t-2)+\delta(t-3)]+\cdots$$
Then I know the Laplace transform of the pairs would be
$$[1-e^{-s}]+[e^{-2s}+e^{-3s}]+\cdots$$
and then group the positive terms ($\sum_{k=0}^{+\infty}e^{-2ks}$) and the negative terms ($-\sum_{k=0}^{+\infty}e^{-(2k+1)s}$) and use the hint to get the final result, but I am unable to get the correct answer. I was wondering if my approach is correct or perhaps there is another way to calculate the Laplace transform?
Thank you in advance.
Best Answer
$$ \mathcal{L} \left\{ \sum_{k=0}^{+\infty}(-1)^{k}\delta (t-k) \right\} = \int_0^\infty \left( \sum_{k=0}^{+\infty}(-1)^{k}\delta (t-k) \right) e^{-st} \, dt = \sum_{k=0}^{+\infty}(-1)^{k} \int_0^\infty \delta (t-k) e^{-st} \, dt \\ = \sum_{k=0}^{+\infty}(-1)^{k} e^{-ks} = \sum_{k=0}^{+\infty}(-e^{-s})^{k} = \frac{1}{1-\left(-e^{-s}\right)} = \frac{1}{1+e^{-s}} = \frac{e^s}{e^s+1} $$