How to find the Laplace transform of $f(t \pm a)$?

Question

I am trying to find the laplace transform of $f(t \pm a)$. I did the integration myself and stopped at this point
$$e^{\pm as}[\int_{\pm a}^{0}e^{-sk}f(k)dk + \int_{0}^{\infty}e^{-sk}f(k)dk]$$
which gives rise to this term
$$e^{\pm as}[\int_{\pm a}^{0}e^{-sk}f(k)dk +F(s)]$$
Where $L[f(t)]=F(s)$ and $ t \pm a =k$
I couldn't figure out the value of this term
$$\int_{\pm a}^{0}e^{-sk}f(k)dk$$
Any help is appreciated.

Answer 1

Don't forget the Heaviside step function, since for $t\le a \implies f(t-a)=0$ $$\mathscr{L}\{f(t-a)u(t-a)\}=\int_0^\infty f(t-a)u(t-a)e^{-st}dt$$ The Heaviside step function, changes the bound of the integral: $$\mathscr{L}\{f(t-a)u(t-a)\}=\int_a^\infty f(t-a)e^{-st}dt$$ Substitute $t-a=u \implies du=dt$ and the bounds change ( for a not at infinity) in the integral $t=a \implies u=0$: $$\mathscr{L}\{f(t-a)u(t-a)\}=\int_0^\infty f(u)e^{-s(u+a)}du$$ $$\mathscr{L}\{f(t-a)u(t-a)\}=e^{-as}\int_0^\infty f(u)e^{-su}du$$ Finally: $$\boxed{\mathscr{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)}$$

Answer 2

Just use the definition.

$$\int_0^{+\infty} e^{-ts}f(t \pm a)\ \text{d}t$$

Using $y = t\pm a$ we easily find

$$\int_0^{+\infty} e^{-s(y \pm a)} f(y)\ \text{d}y = e^{\pm sa} \hat{f}(y)$$

This is a well known property, called frequency shifting. Take a look at the "properties" section here: https://en.wikipedia.org/wiki/Laplace_transform