Laplace transform list of properties

Consider the Laplace transform of a original function f.

$$\mathcal{L}(f(t))(s)=\int_{0}^{+\infty}f(t)e^{-st}dt=F(s)$$

We give some useful property if this transform.

Change of scale.
Let $ a>0 $ then g(t)=f(at) is function which admits Laplace transform and

$$\mathcal{L}(g(t))(s)=\int_{0}^{+\infty}f(at)e^{-st}dt=\frac{1}{a}\int_{0}^{+\infty}f(\tau)e^{-\frac{s}{a}\tau}d\tau=\frac{1}{a}F(\frac{s}{a})$$

Multiply by exponential
Let $ \lambda\in C $ then
$ g(t)=e^{\lambda t}f(t) $ has Laplace transform

$$\mathcal{L}(g(t))(s)=\mathcal{L}(e^{\lambda t}f(t))(s)=\int_{0}^{+\infty}e^{\lambda t}f(t)e^{-st}dt=\int_{0}^{+\infty}f(t)e^{-(s-\lambda )t}dt=F(s-\lambda)$$

Product with t variable.
If f is an original function, $ f\in\mathcal{O}_{\sigma_0} $ then for $ \forall \epsilon>0,g_k(t)=t^kf(t)\in\mathcal{O}_{\sigma_0+\epsilon} $ and

$$\mathcal{L}(tf(t))(s)=\int_{0}^{+\infty}tf(t)e^{-st}dt=-F'(s)$$
$$\mathcal{L}(t^2f(t))(s)=\int_{0}^{+\infty}t^2f(t)e^{-st}dt=(-1)^2F"(s)$$
$$\mathcal{L}(t^nf(t))(s)=\int_{0}^{+\infty}t^nf(t)e^{-st}dt=(-1)^nF^{(n)}(s)$$

Laplace transform for derivative
If f and f' are original function then

$$\mathcal{L}(f'(t))(s)=\int_{0}^{+\infty}f'(t)e^{-st}dt=f(t)e^{-st}|_0^{+\infty}+s\int_0^{+\infty}f(t)e^{-st}dt=sF(s)-f(0+)$$

Where

$$f(0+)=\lim_{t\to 0, t>0}f(t)$$

If higher order derivatives exists and are originals,

$$\mathcal{L}(f"(t))(s)=\int_{0}^{+\infty}f"(t)e^{-st}dt=s^2F(s)-f(0+)s-f'(0+)$$
$$\mathcal{L}(f^{(n)}(t))(s)=\int_{0}^{+\infty}f^{(n)}(t)e^{-st}dt=s^nF(s)-f(0+)s^{n-1}-f'(0+)s^{n-2}-....-f^{(n-1)}(0+)=$$

Convolution of originals.
Let f,g be original function.
Then

$$f*g(t)=\int_{0}^{+\infty}f(\tau)g(t-\tau)d\tau=\int_{0}^{t}f(\tau)g(t-\tau)d\tau$$

is an original and

$$\mathcal{L}(f*g(t))(s)=\int_{0}^{+\infty}f*g(t)e^{-st}dt=\int_{0}^{+\infty}\int_{0}^{t}f(\tau)g(t-\tau)d\tau e^{-st}dt=$$
$$=\int_{0}^{+\infty}f(\tau)e^{-s\tau}(\int_{\tau}^{+\infty}g(t-\tau) e^{-s(t-\tau)}dt)d\tau=\int_{0}^{+\infty}f(\tau)e^{-s\tau}d\tau\int_{\tau}^{+\infty}g(t-\tau)e^{-s(t-\tau)}dt=$$
$$=\int_{0}^{+\infty}f(\tau)e^{-s\tau}d\tau\int_{0}^{+\infty}g(t')e^{-st'}dt'=F(s)G(s)$$

Let f be a periodic function of period T >0, which take value zero on the negative real axis.
Suppose that

$$f_0(t)=f(t)(h(t-T)-h(t))$$

is integrable.
Then Laplace transform of f

$$\mathcal{L}(f(t)(s)=\int_{0}^{+\infty}f(t)e^{-st}dt=\frac{1}{1-e^{-sT}}\int_{0}^{T}f(t)e^{-st}dt=\mathcal{L}(f_0(t))(s)$$
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