How to find inverse Laplace transform of $\frac{\sqrt{s} \sqrt{a+s}}{b+s}$

Question

How to find inverse Laplace transform of

$$\frac{\sqrt{s} \sqrt{a+s}}{b+s}$$

where $a$ and $b$ constants, ${a,b}>0$

I tried to solve it, but I couldn't.

EDITED.

Numerical solution for $a=1,b=1$. Plot:

Answer 1

One method of consideration may be applied to the following. By using the series expansion $$\sqrt{1-x} = 1 - \sum_{k=1}^{\infty} \frac{(2k-2)! \, x^{k}}{2^{2k-1} \, k! \, (k-1)!}.$$ Now the fraction in question becomes: \begin{align} \frac{\sqrt{s \, (s+a)}}{s+b} &= \frac{\sqrt{\left(s+ \frac{a}{2}\right)^{2} - \frac{a^{2}}{4}}}{s+b} = \frac{s + \frac{a}{2}}{s+b} \cdot \sqrt{1 - \frac{a^{2}}{4 \, \left(s + \frac{a}{2}\right)^{2}}} \\ &= \frac{s + \frac{a}{2}}{s+b} - \sum_{k=1}^{\infty} \frac{(2k-2)! \, \left(\frac{a}{2}\right)^{2k}}{2^{2k-1} \, k! \, (k-1)!} \cdot \frac{1}{(s+b) \, \left(s + \frac{a}{2}\right)^{2k-1}}. \end{align} By using the convolution theorem it is developed into: \begin{align} \frac{(2k-2)!}{(s+b) \, \left(s + \frac{a}{2}\right)^{2k-1}} &\Doteq \int_{0}^{t} e^{-b (t-u)} \, u^{2k-2} \, e^{- \frac{a u}{2}} \, du \\ &\Doteq \frac{e^{-bt}}{\left(\frac{a}{2} - b\right)^{2k-1}} \, \int_{0}^{\left(\frac{a}{2} - b\right) \, t} e^{-x} \, x^{2k-2} \, dx \\ &\Doteq \frac{e^{-bt}}{\left(\frac{a}{2} -b\right)^{2k-1}} \, \gamma\left(2k-1, \left(\frac{a}{2} - b\right) \, t \right), \end{align} where $\gamma(a,x)$ is the incomplete Gamma function. Putting all the components together yields $$\frac{\sqrt{s \, (s+a)}}{s+b} \Doteq \delta(t) + \left(\frac{a}{2} - b\right) \, e^{-b t} \, \left[1 - 2 \, \sum_{k=1}^{\infty} \frac{\left(\frac{a}{2 \, (a-2b)}\right)^{2k} \, \gamma\left(2k-1, \left(\frac{a}{2} -b\right) \, t \right) }{ k! \, (k-1)!} \right]. $$

Notice that if $b = \frac{a}{2}$ then this reduces to $$\frac{\sqrt{s \, (s+a)}}{s+\frac{a}{2}} \Doteq \delta(t) - \frac{a^{2} \, t}{8} \, e^{- \frac{a \, t}{2}} \, {}_{1}F_{2}\left(\frac{1}{2}; \frac{3}{2} , 2; \frac{a^{2} \, t^{2}}{16} \right).$$