Let $f:[0,\infty[\to\mathbb {R}$ be the periodic function with period $T=1$ defined on $[0,\infty[$ by $$f(t)=t,\,\,\,\,\,\text{pour}\,\,0\leq t<1.$$
The book I am reading used the following formula of Laplace transform of a periodic function $$\mathcal{L}\left(f\right)=\frac{{\displaystyle \int_{0}^{T}f(t)e^{-st}dt}}{1-e^{-sT}}$$
to find that $$\mathcal{L}\left(f\right)=\frac{1}{s^2}-\frac{e^{-s}}{s(1-e^{-s})}.$$
I wanted to find it using another method so I used the Laplace transform of the derivative: $$f'(t)=1-\sum_{n=1}^\infty \delta(t-n)$$ thus I applied L.T: $$\mathcal{L}(f')=\mathcal{L}(1)-\sum_{n=1}^\infty \mathcal{L}(\delta(t-n))$$ $$s\mathcal{L}(f)=\frac{1}{s}-\sum_{n=1}^\infty e^{-ns}=\frac{1}{s}-\frac{e^{-s}}{1-e^{-s}}$$ $$\mathcal{L}\left(f\right)=\frac{1}{s^2}-\frac{e^{-s}}{s(1-e^{-s})}.$$
I got the same result, but
(1) I am just wondering if the operations I did like the formula $\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$ are correct even when using a distributional derivative $f'$ like in this case.
(2) I am also wondering if I am able to apply the linearity of $\mathcal{L}$ to the infinite sum: $$\mathcal{L}\left(\sum_{n=1}^\infty \delta(t-n)\right)=\sum_{n=1}^\infty \mathcal{L}\left(\delta(t-n)\right)$$
Think in term of the bilateral Laplace transform, then, for $g e^{-\sigma t},g e^{-\sigma_2 t}$ a tempered distribution, for $\Re(s)\in (\sigma,\sigma_2)$.. $$\mathcal{L}(g') = s \mathcal{L}(g)$$ Here you are looking at $g(t)=\sum_{n\ge 0} (t-n) 1_{t-n\in [0,1)}$ so that $g'(t)= 1_{t > 0}-\sum_{n\ge 1}\delta(t-n)$, $\mathcal{L}(g') = \frac1s -\sum_{n\ge 1}e^{-sn}$