I am trying to inverse Laplace transform of $\frac{\sqrt{1+s}}{\sqrt{s}+a}$ where a is a complex constant. Here are the three facts I get:
$$ \mathcal{L}_{s}^{-1}\left[\frac{1}{a+k s^{\alpha}}\right](t)=\frac{t^{-1+\alpha} E_{\alpha, \alpha}\left(-\frac{a t^{\alpha}}{k}\right)}{k} $$ where $E_{\alpha, \alpha}\left(-\frac{a t^{\alpha}}{k}\right)$ is the generalized Mittag-Leffler function.
$$ \mathcal{L}_{s}^{-1}\left[\sqrt{s+a}\right](t) = -\frac{e^{-a t}}{2 \sqrt{\pi t^{3}}} $$
$$ \mathcal{L}^{-1}[\sqrt{1+2 a / s}-1](t)=a e^{-a t}\left(I_{0}(a t)+I_{1}(a t)\right) $$ where $I_{\nu}$ is the modified bessel function of first kind.
I do not know how to use these identities to do this $\frac{\sqrt{1+s}}{\sqrt{s}+a}$? Can someone help me? thank you!
If $a=0$ you can use $(3)$ on $\sqrt{1+1/s}-1+1$. Otherwise you can try to use convolution theorem along with $(1)$ and $(2)$ on $\frac{1}{2a}(\sqrt{s+1})(\dfrac{1}{\dfrac{1}{2a}\sqrt{s}+\dfrac{1}{2}})$. However, I'm not sure how to evaluate the resulting integral. I'm stuck at evaluating the definite integral of the function in $(2)$, but there might be some complex analysis techniques that can solve this.