Inverse Laplace Tranform of $\frac{s}{s^2+w_0^2} e^{-x\sqrt{s/\alpha}}$ I want to calculate the following Inverse Laplace Transform:
$$ \mathcal L^{-1} \left\{ \dfrac{s}{s^2+w_0^2} e^{-x \sqrt{\dfrac{s}{\alpha}}} \right\}$$
I was not able to find such expression in any table, so I decided to go with the integral, that reduces to the calculation of the residues of the function
$$F(s) = \dfrac{s}{s^2+w_0^2} e^{-x \sqrt{\dfrac{s}{\alpha}} + st}$$
$F(s)$ has two simple poles at $s=\pm w_0 i$. Then:
$$\text{Res}(F(s),w_0 i) = \frac{1}{2} e^{-\sqrt{\dfrac{w_0}{\alpha}} x \left( \dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2} i \right) + w_0 i t}$$
$$\text{Res}(F(s),- w_0 i) = \frac{1}{2} e^{-\sqrt{\dfrac{w_0}{\alpha}} x \left( \dfrac{1}{\sqrt 2} - \dfrac{1}{\sqrt 2} i \right) - w_0 i t}$$
Then:
$$ \mathcal L^{-1} \left\{ \dfrac{s}{s^2+w_0^2} e^{-x \sqrt{\dfrac{s}{\alpha}}} \right\} =  \text{Res}(F(s),w_0 i) + \text{Res}(F(s),-w_0 i) $$
$$= e^{-\sqrt{\dfrac{w_0}{2\alpha}} x} \cos \left( \sqrt{\dfrac{w_0}{2\alpha}} + w_0 t \right)$$
However, this does not seem to be the correct expression for the inverse Laplace Transform. If I apply the inital value theorem:
$$ \lim_{s\to\infty} \left[ s \dfrac{s}{s^2+w_0^2} e^{-x \sqrt{\dfrac{s}{\alpha}}}  \right] = 0$$
Whereas
$$ \lim_{t\to 0} \left[ e^{-\sqrt{\dfrac{w_0}{2\alpha}} x} \cos \left( \sqrt{\dfrac{w_0}{2\alpha}} + w_0 t \right) \right] = e^{-\sqrt{\dfrac{w_0}{2\alpha}} x} \cos \left( \sqrt{\dfrac{w_0}{2\alpha}} \right) \neq 0$$
¿Where do I go wrong? ¿What is the correct way to calculate this Inverse Laplace transform and, if you may, what is the actual result?
Thanks in advance.
 A: Well, we can use the convolution theorem:
$$\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{s}^2+\omega_0^2}\cdot\exp\left\{-x\cdot\sqrt{\frac{\text{s}}{\alpha}}\right\}\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{s}^2+\omega_0^2}\right]_{\left(t\right)}\space*\space\mathscr{L}_\text{s}^{-1}\left[\exp\left\{-x\cdot\sqrt{\frac{\text{s}}{\alpha}}\right\}\right]_{\left(t\right)}$$
Now, we know that:
$$\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{s}^2+\omega_0^2}\right]_{\left(t\right)}=\cos\left(\omega_0t\right)$$
And:
$$\mathscr{L}_\text{s}^{-1}\left[\exp\left\{-x\cdot\sqrt{\frac{\text{s}}{\alpha}}\right\}\right]_{\left(t\right)}=\sum_{\text{n}=0}^\infty\frac{1}{\text{n}!}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(-x\cdot\sqrt{\frac{\text{s}}{\alpha}}\right)^\text{k}\right]_{\left(t\right)}$$
We can use:
$$\left(-x\cdot\sqrt{\frac{\text{s}}{\alpha}}\right)^\text{k}=\left(-1\right)^\text{k}\cdot x^\text{k}\cdot\frac{\text{s}^\frac{\text{k}}{2}}{\alpha^\frac{\text{k}}{2}}$$
