# How to find the Laplace transform of f(t±a)?

How to find the Laplace transform of $f\left(t±a\right)$?
I am trying to find the laplace transform of $f\left(t±a\right)$. I did the integration myself and stopped at this point
${e}^{±as}\left[{\int }_{±a}^{0}{e}^{-sk}f\left(k\right)dk+{\int }_{0}^{\mathrm{\infty }}{e}^{-sk}f\left(k\right)dk\right]$
which gives rise to this term
${e}^{±as}\left[{\int }_{±a}^{0}{e}^{-sk}f\left(k\right)dk+F\left(s\right)\right]$
Where $L\left[f\left(t\right)\right]=F\left(s\right)$ and $t±a=k$
I couldn't figure out the value of this term
${\int }_{±a}^{0}{e}^{-sk}f\left(k\right)dk$
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Peugeota2p
Heaviside step function, since for $t\le a\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(t-a\right)=0$
$\mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(t-a\right)u\left(t-a\right){e}^{-st}dt$
The Heaviside step function, changes the bound of the integral:
$\mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}={\int }_{a}^{\mathrm{\infty }}f\left(t-a\right){e}^{-st}dt$
Substitute $t-a=u\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}du=dt$ and the bounds change ( for a not at infinity) in the integral $t=a\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}u=0$:
$\mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(u\right){e}^{-s\left(u+a\right)}du$
$\mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}={e}^{-as}{\int }_{0}^{\mathrm{\infty }}f\left(u\right){e}^{-su}du$
Finally:
$\overline{)\mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}={e}^{-as}F\left(s\right)}$

Hrefnui9
Just use the definition.

Using $y=t±a$ we easily find

This is a well known property, called frequency shifting.