Inverse Laplace transform of $\frac{e^{-a\sqrt{s}}}{(s-r)(\sqrt{s}-b)}$

Question

I'm trying to find an expression for the inverse Laplace transform of $\frac{e^{-a\sqrt{s}}}{(s-r)(\sqrt{s}-b)}$. Partial fraction decomposition gives $\frac{e^{-a\sqrt{s}}}{(b^2-r)(\sqrt{s}-b)} - \frac{(b + \sqrt{s})e^{-a\sqrt{s}}}{(b^2-r)(s-b)}$ but that didn't seem to help much.

Answer 1

Now, let's start with the proper example:

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. Update:

After arduous and lengthy calculations using Maple and Mathematica finally succeeded:

$\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\cdot\sqrt{\text{s}}}}{\left(\text{s}-\text{r}\right)\cdot\left(\sqrt{\text{s}}-\text{b}\right)}\right]_{\left(t\right)}=\frac{e^{-a \sqrt{r}} \left(2 b \text{erfc}\left(\frac{a-2 b t}{2 \sqrt{t}}\right) e^{-a b+a \sqrt{r}+b^2 t}+e^{r t} \left(\left(-b-\sqrt{r}\right) \text{erfc}\left(\frac{a-2 \sqrt{r} t}{2 \sqrt{t}}\right)+e^{2 a \sqrt{r}} \left(\sqrt{r}-b\right) \text{erfc}\left(\frac{a}{2 \sqrt{t}}+\sqrt{r t}\right)\right)\right)}{2 \left(b^2-r\right)}$

Check numerics:

Answer 2

Use the convolution theorem:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\cdot\sqrt{\text{s}}}}{\left(\text{s}-\text{r}\right)\cdot\left(\sqrt{\text{s}}-\text{b}\right)}\right]_{\left(t\right)}=\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}-\text{r}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\cdot\sqrt{\text{s}}}}{\sqrt{\text{s}}-\text{b}}\right]_{\left(t\right)}=$$ $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}-\text{r}}\right]_{\left(t\right)}*\left\{\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\sqrt{\text{s}}-\text{b}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[e^{-\text{a}\cdot\sqrt{\text{s}}}\right]_{\left(t\right)}\right\}\tag1$$

Now, using:

1. $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}-\text{r}}\right]_{\left(t\right)}=e^{\text{r}\cdot t}\tag2$$
2. When $\text{b}<0$: $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\sqrt{\text{s}}-\text{b}}\right]_{\left(t\right)}=\frac{1}{\sqrt{\pi}\cdot\sqrt{t}}+\text{b}\cdot e^{\text{b}^2\cdot t}\cdot\text{erfc}\left(-\text{b}\cdot\sqrt{t}\right)\tag3$$
3. When $\text{a}>0$: $$\mathcal{L}_\text{s}^{-1}\left[e^{-\text{a}\cdot\sqrt{\text{s}}}\right]_{\left(t\right)}=\frac{\text{a}\cdot e^{-\frac{\text{a}^2}{4\cdot t}}}{2\cdot\sqrt{\pi}\cdot t^{\frac{3}{2}}}\tag4$$

So:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\cdot\sqrt{\text{s}}}}{\left(\text{s}-\text{r}\right)\cdot\left(\sqrt{\text{s}}-\text{b}\right)}\right]_{\left(t\right)}=e^{\text{r}\cdot t}*\left\{\left(\frac{1}{\sqrt{\pi}\cdot\sqrt{t}}+\text{b}\cdot e^{\text{b}^2\cdot t}\cdot\text{erfc}\left(-\text{b}\cdot\sqrt{t}\right)\right)*\frac{\text{a}\cdot e^{-\frac{\text{a}^2}{4\cdot t}}}{2\cdot\sqrt{\pi}\cdot t^{\frac{3}{2}}}\right\}$$