Laplace transform of a ramp function with a $2$ second delay Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.
$$\int_{2}^\infty t e^{-st} \, \mathrm{d}t = \frac{-te^{-st}}{s}\bigg|_{2}^\infty + \int_{2}^\infty \frac{e^{-st}}{s} \, \mathrm{d}t = \frac{2e^{-2s}}{s}+\frac{e^{-2s}}{s^2}$$
This is what I get. Where am I going wrong?
 A: The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ \mathcal{L} [f (t- 2) u(t-2)] = \int_0^\infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + \int_2^\infty f(t-2)\, e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity. 
To integrate this we make a change of variable were $\tau = t - 2$, hence $t = \tau + 2$ and $dt = d\tau$. Note that the lower limit changes because when $t=2$, $\tau = 2 - 2 = 0$, so that
$$ = \int_0^\infty f(\tau) e^{-s(\tau + 2)} d\tau $$
$$ = e^{-2s} \int_0^\infty f(\tau) e^{-s\tau} d\tau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ \mathcal{L} [f (t- 2) u(t-2)] = \frac{e^{-2s}}{s^2} $$
