Inverse Laplace Transform of $\frac{1}{\sqrt{s+a}+\sqrt{s+b}}$

Question

I need to calculate the inverse laplace of: $$F(s)=[\frac{1}{\sqrt{s+a}+\sqrt{s+b}}] \qquad \qquad (s>-a\quad ;\quad s>-b;\quad a\neq b) $$

Answer 1

Note that by multiplying the numerator and the denominator of $F(s)$ by $\sqrt{s+a}-\sqrt{s+b}$ you get

$$F(s)=\frac{\sqrt{s+a}-\sqrt{s+b}}{a-b}\tag{1}$$

The inverse Laplace transform of $\sqrt{s+a}$ can be found in tables:

$$\sqrt{s+a}\leftrightarrow -\frac{e^{-at}}{2 \sqrt{\pi t^3}}\tag{2}$$

With $(1)$ and $(2)$ we get for the inverse transform of $F(s)$

$$F(s)\leftrightarrow \frac{e^{-bt}-e^{-at}}{2(a-b) \sqrt{\pi t^3}}\tag{3}$$