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For $a > 0$, is there any known representation of the Laplace transform of $f(t+a)$ in terms of the Laplace Transform of $f(t) $
Note: In my application, $f(t)$ is not a periodic function and the functional form of $f(t)$ is not actually known a-priori, because I have to couple it to another set of equations.
Let $s = \sigma + j\omega$, the inverse Laplace transform of $f(t+a)$ is given by $$f(t+a) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{s(t+a)} \mathrm{d}s = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{sa}e^{st} \mathrm{d}s.$$
Hence the bilateral Laplace transform of $f(t+a)$ is $F(s)e^{sa}$ where $F(s)$ is the Laplace transform of $f(t)$. For the unilateral case, see Matt L. answer.
This is sometimes called the shifting theorem. See Theorem 12.16 here.
You have to be clear about which flavor of the Laplace transform you're talking about. If you consider the bilateral Laplace transform
$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt\tag{1}$$
then the relationship
$$\mathcal{L}\{f(t+a)\}=e^{as}F(s)\tag{2}$$
clearly holds, also for $a>0$.
However, if you consider the unilateral Laplace transform
$$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt\tag{3}$$
then for $a>0$
$$\mathcal{L}\{f(t+a)\}=e^{as}\int_a^{\infty}f(t)e^{-st}dt\tag{4}$$
which is generally not equal to $e^{as}F(s)$ (unless $f(t)=0$ in $[0,a]$).
In your example $f(t)=t^2$ you implicitly use the unilateral Laplace transform, so $(2)$ doesn't hold.
