Demonstration with Laplace

Question

How can I demonstrate this? If $F(t)$ is a periodic function with a period of $T>0$, then

$$ \mathcal{L}\{F(t)\} = \frac{\int\limits_0^T e^{-st} F(t)\operatorname d\!t}{1-e^{-sT}}\operatorname d\!t$$

Any help will be appreciated.

Answer 1

Hint:

$$\int_0^\infty e^{-st} F(t)\,dt = \sum_{k=0}^\infty \int_{kT}^{(k+1)T} e^{-st}F(t)\,dt = \sum_{k=0}^\infty \int_0^T e^{-s(t+kT)}F(t+kT)\,dt.$$

Use the periodicity of $F$ and the formula for the sum of a geometric series.

Answer 2

You can also use the shifting property of the Laplace transform to obtain the result. Consider the Laplace transform of the first period of the signal and add the other periods using the shifting property:

$$\mathcal{L}\{F(t)\}=\int_0^TF(t)e^{-st}dt\cdot\left(1+e^{-sT}+e^{-2sT}+\ldots\right)=\int_0^TF(t)e^{-st}dt\cdot\sum_{n=0}^{\infty}e^{-nsT}=\\ =\frac{1}{1-e^{-sT}}\int_0^TF(t)e^{-st}dt$$