How to obtain Laplace transform of {f(t-a)U(t-b)}

Question

$f(t)=g(t-10)U(t-15)-g(t-10)U(t-20)$

The above $f(t)$ contains terms of the form $g(t-a)U(t-b)$, where $a$ doesn't equal $b$. Describe the form that $L\{f(t-a)U(t-b)\}$ takes. [Hint: The formula for $L\{g(t)U(t-a)\}=e^{-as}L\{g(t+a)\}$

Answer 1

$$Lf(t) = \int_0^\infty e^{-st}f(t)dt$$

$$L\{g(t-a)U(t-b)\}= \int_0^\infty e^{-st}(g(t-a)U(t-b))dt\\ = \int_b^\infty e^{-st}g(t-a)dt = \int_{b-a}^\infty e^{-s(v+a)}g(v)dv = e^{-sa}\int_{0}^\infty e^{-sv}g(v)U(b-a)dv = e^{-sa}\left(Lg(t)-\int_{0}^{b-a} e^{-sv}g(v)dv\right) $$

Thus $$L\{g(t-10)U(t-15)-g(t-10)U(t-20)\} =\\ e^{-10s}\left(Lg(t)-\int_{0}^{5} e^{-sv}g(v)dv\right) -e^{-10s}\left(Lg(t)-\int_{0}^{10} e^{-sv}g(v)dv\right)=\\ e^{-10s}\left(\int_{0}^{10} e^{-sv}g(v)dv-\int_{0}^{5} e^{-sv}g(v)dv\right)=\\ e^{-10s}\int_{5}^{10} e^{-sv}g(v)dv $$

Answer 2

HINT: $L\{ g(t-10)U(t-15)\}=e^{-15s}g(t-10+15)=e^{-15s}L\{g(t+5)\}$

and $L\{ g(t-10)U(t-20)\}=e^{-20s}g(t-10+20)=e^{-20s}L\{g(t+10)\}$

Alternatively, the given function is:

$f(t)=g(t-10), 15\le t\le 20$

Then, using standard definition $L\{f(t)\}=\int_{15}^{20} e^{-st}g(t-10)dt$