Laplace transform of periodic functions

Let $f(t)$ be periodic with the positive period (http://planetmath.org/PeriodicFunctions) $p$.  Denote by $H(t)$ the Heaviside step function.  If now

$$g(t):=f(t)H(t)-f(t-p)H(t-p),$$

then it follows

(1)

By the parent entry (http://planetmath.org/DelayTheorem), the Laplace transform of $g$ is

$$G(s)=F(s)-e^{-ps}F(s),$$

whence

$$F(s)=\frac{G(s)}{1-e^{-ps}}=\frac{1}{1-e^{-ps}}{\int }_0^{\mathrm{\infty }}{e}^{-st}g(t)𝑑t=\frac{1}{1-e^{-ps}}{\int }_0^p{e}^{-st}f(t)𝑑t.$$

Thus we have the rule

$ℒ\{f(t)\}={\displaystyle \frac{1}{1-e^{-ps}}}{\displaystyle {\int }_0^p}{e}^{-st}f(t)𝑑t\mathit{\hspace{1em}\hspace{1em}}(\text{period }p).$(2)

On the contrary, if $f(t)$ is antiperiodic with positive antiperiod $p$, then the function

$$g(t):=f(t)H(t)+f(t-p)H(t-p)$$

also has the property (1).  Analogically with the preceding procedure, one may derive the rule

$ℒ\{f(t)\}={\displaystyle \frac{1}{1+e^{-ps}}}{\displaystyle {\int }_0^p}{e}^{-st}f(t)𝑑t\mathit{\hspace{1em}\hspace{1em}}(\text{antiperiod }p).$(3)