The piecewise function is
$f(t)=$
\begin{cases} 0 & t\leq 0 \\ \sqrt{t} & 0 \leq t\leq 4 \\ 2 \cos(t-4) & 4\leq t <4+\pi/2\\ 0 & t >4+\pi/2\\ \end{cases}
I have expressed it as:
$f(t)=\sqrt{t}[H(t)-H(t-4)]+2 \cos(t-4)[H(t-4)-H(t-(4+\pi/2)]$
However, this doesn't seem to simplify into something where the delay formula can be used. I know that
$L(\sqrt{t})=\dfrac{\sqrt\pi}{s^{3/2}}$ and $L(\cos(at-b)=\dfrac{s\cdot\cos(b)+\sin(b)}{s^2+a^2}$
but I'm not sure how to proceed with this question.
edit:
I have now written this as
$f(t)=\int _0^4 \sqrt{t} \cdot e^{-st} dt + 2\int _4^{4+\pi/2} \cos(t-4) \cdot e^{-st} dt$
is this correct?
I hate to provide such a simple and unenlightening answer, but to your last question: yes.
First note that, $$ e^{-st}f(t) = \begin{cases} 0 & t < 0 \\ e^{-st}\sqrt{t} & 0 \leq t < 4 \\ 2e^{-st}\cos(t-4) & 4 \leq t < 4+\frac{\pi}{2} \\ 0 & t \geq 4+\frac{\pi}{2} \\ \end{cases} $$
Then using the definition of the Laplace transform and some basic properties of the integral operator, \begin{align*} L(f(t)) &:= \int_{-\infty}^{\infty}e^{-st}f(t)dt \\ \\ &= \int_{0}^{4}e^{-st}\sqrt{t}dt + 2\int_{4}^{4+\frac{\pi}{2}}e^{-st}\cos(t-4)dt \\ \\ &= \frac{\sqrt{\pi}\text{erf}(2\sqrt{s})}{2s^{3/2}} - \frac{2e^{-4s}}{s} + \frac{2(se^{-4s}+e^{-\frac{1}{2}(8+\pi)s})}{s^2+1} \end{align*}
where the left integral (for the $\sqrt{t}$ term) is not known in closed form so we make use of the error function, $$ \text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2}dt $$
but the right integral (for the $\cos$ term) can be carried out with integration by parts.
I'm curious if there is a more elegant solution to this as well, but as far as I can tell, that's as nice as it gets.
Did this problem come from a textbook? I'm curious about the context. In many applications, we are concerned with the Laplace transform of a sum of "forcing functions" that may "switch on" at various times. For example, if instead your problem had, $$ f(t) := \begin{cases} 0 & t < 0 \\ \sqrt{t} & 0 \leq t < 4 \\ \sqrt{t} + 2\cos(t-4) & 4 \leq t < 4+\frac{\pi}{2} \\ \sqrt{t} & t \geq 4+\frac{\pi}{2} \\ \end{cases} $$ it would be much simpler to evaluate. In the context of linear ordinary differential equations, the above would represent a square-root input "activating" at time 0 and a quarter-period sinusoidal pulse input at time 4.
I mention this because if the piecewise function you gave was an interpretation of something else, perhaps a mistake was made in the interpretation, leading to this otherwise challenging problem.
One other thing that may make it possible is multiplying $f(t)$ by some function $g(t)$ before taking the laplace transform, where there is some known relationship between $L(f(t))$ and $L(g(t)f(t))$ that can be exploited after the fact to "undo" our application of $g(t)$. For example, $$L(tf(t)) = -\frac{d}{ds}L(f(t))$$ Unfortunately I don't see a $g(t)$ that exactly works for you.
The last thing I can think of that may be worth exploring is the convolution theorem.
If any of these methods were to work and give you an answer in terms of simple functions, it would mean that Wolfram was wrong and that, $$\int_{a}^{b}e^{-st}\sqrt{t}dt$$ does have a closed-form solution. This is why I am skeptical.