<s>[INST] Laplace transform $\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)$ How to find Laplace transform ?:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s) $$
I have been stuck on this problem for quite a bit. I only found in tables:
$$\mathcal{L}_x\left[\frac{1}{(x+8) \sqrt{x+8}}\right](s)=\frac{1}{\sqrt{2}}-2 e^{8 s} \sqrt{\pi } \sqrt{s} \text{erfc}\left(2 \sqrt{2} \sqrt{s}\right)$$
My attempt:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)
=\mathcal{L}_x\left[\int_0^{\infty } \frac{\exp (-t (x+9))}{\sqrt{x+8}} \, dt\right](s)=\int_0^{\infty }
   \mathcal{L}_x\left[\frac{\exp (-t (x+9))}{\sqrt{x+8}}\right](s) \, dt=\int_0^{\infty } \frac{e^{8s-t} \sqrt{\pi } \text{erfc}\left(2 \sqrt{2} \sqrt{s+t}\right)}{\sqrt{s+t}}
   \, dt$$
or:
$$\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)=\int_0^{\infty } -\frac{2 e^{9 s+t^2} \text{Ei}\left(-9 \left(s+t^2\right)\right)}{\sqrt{\pi }} \, dt$$
 [/INST] The integral you need to compute is
$$
I(s)=\int_0^\infty dx\frac{e^{-sx}}{(x+9)\sqrt{x+8}}\ .
$$
Call $\sqrt{x+8}=t$. Then $x=t^2-8$ and $dx=2t dt$.
So
$$
I(s)=\int_{2\sqrt{2}}^\infty 2t dt\frac{e^{-s(t^2-8)}}{t(t^2+1)}=2e^{8s}\int_{2\sqrt{2}}^\infty dt\frac{e^{-s t^2}}{t^2+1}\ .
$$
This can be rewritten as
$$
I(s)=2e^{8s}\left[\int_0^\infty dt\frac{e^{-s t^2}}{t^2+1}-\int_0^{2\sqrt{2}}dt\frac{e^{-s t^2}}{t^2+1}\right]=2e^{8s}\left[\frac{1}{2} \pi  e^s \text{erfc}\left(\sqrt{s}\right)-2\pi e^s \frac{1}{2\pi}\int_0^{2\sqrt{2}}dt\frac{e^{-s (t^2+1)}}{t^2+1}\right]
=2e^{9s}\pi\left[\frac{1}{2}  \text{erfc}\left(\sqrt{s}\right)-2 T(\sqrt{2s},2\sqrt{2})\right]\ ,
$$
where $T(x,a)$ is the Owen's T-function (https://reference.wolfram.com/legacy/v9/ref/OwenT.html).
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