Inverse Laplace Transform for $ \frac{1}{p} \cdot \frac{1}{\sqrt{p + b}}\cdot e^{-x \sqrt{p+b}} $ I am looking for the analytical formula of the Laplace original for 
$ \frac{1}{p} \cdot \frac{1}{\sqrt{p + b}}\cdot e^{-x \sqrt{p+b}}, $ where $ b, x  \ge 0. $ I know that it exists, since years ago I myself derived it, but did not write down how, and now I cannot repeat it any longer. The Laplace image arises in many problems in heat transfer, diffusion and similar domains, so I am sure that many people have already solved it before. I would be thankful for a solution idea or, even better, a reference where it can be found.
 A: The Inverse Laplace Transform (ILT) of the function
$$F(p) = \frac1{p \sqrt{p+b}} e^{-\xi \sqrt{p+b}} $$
$b, \xi \gt 0$, is given by the complex integral, for $t > 0$
$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \, F(p) e^{p t} $$
This complex integral may be evaluated using the following contour integral:
$$\oint_C dz \, \frac1{z \sqrt{z+b}} e^{-\xi \sqrt{z+b}} e^{z t} $$
where $C$ is the contour pictured below, where the small loop is centered at $z=-b$ and has radius $\epsilon$; the large loop is centered at the origin and has radius $R$.  .

We may thus write out the contour integral as follows:
$$\int_{c-\sqrt{R^2-c^2}}^{c+\sqrt{R^2-c^2}} dp \, \frac1{p \sqrt{p+b}} e^{-\xi \sqrt{p+b}} e^{p t} + i \int_{\arcsin{(c/r)}}^{\pi} d\theta \, \frac{e^{-\xi \sqrt{R e^{i \theta}+b}}}{\sqrt{R e^{i \theta}+b}} e^{R t e^{i \theta}} \\ + \int_R^{b-\epsilon} \frac{dx}{x} \frac{e^{-i \xi \sqrt{x-b}}}{i \sqrt{x-b}} e^{-x t} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, \frac{e^{-\xi \sqrt{\epsilon} e^{i \phi/2}}}{\sqrt{\epsilon} e^{i \phi/2}} e^{(-b + \epsilon e^{i \phi}) t} \\ + \int_{b-\epsilon}^R \frac{dx}{x} \frac{e^{i \xi \sqrt{x-b}}}{-i \sqrt{x-b}} e^{-x t} + i \int_{-\pi}^{-\arcsin{(c/r)}} d\theta \, \frac{e^{-\xi \sqrt{R e^{i \theta}+b}}}{\sqrt{R e^{i \theta}+b}} e^{R t e^{i \theta}}$$
As $R \to \infty$, the second and sixth integrals $\to 0$; as $\epsilon \to 0$, the fourth integral $\to 0$.  By the residue theorem, the contour integral is equal to $i 2 \pi$ times the residue of the integrand at $z=0$, which is $e^{-\xi \sqrt{b}}/{\sqrt{b}}$.  Putting everything together, we get an expression for the ILT:
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \, \, \frac1{p \sqrt{p+b}} e^{-\xi \sqrt{p+b}} e^{p t} = \frac{e^{-\xi \sqrt{b}}}{\sqrt{b}} - \frac1{\pi} \int_0^{\infty} \frac{dx}{x+b} \frac{\cos{(\xi \sqrt{x})}}{\sqrt{x}} e^{-x t} $$
The integral on the RHS may be evaluated by subbing $x=u^2$ and expressing in terms of a Fourier transform.
$$\frac1{\pi} \int_0^{\infty} \frac{dx}{x+b} \frac{\cos{(\xi \sqrt{x})}}{\sqrt{x}} e^{-x t} = \frac1{\pi} \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{u^2+b} e^{i \xi u} $$
To evaluate the Fourier transform, we use the convolution theorem: the Fourier transform of the product of two functions is the convolution of their Fourier transforms.  Thus, we may write
$$\begin{align} \frac1{\pi} \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{u^2+b} e^{i \xi u} &= \frac1{\pi} \frac1{2 \pi} \int_{-\infty}^{\infty} d\xi' \, \frac{\pi}{\sqrt{b}} e^{-\sqrt{b} |\xi'|} \sqrt{\frac{\pi}{t}} e^{-(\xi-\xi')^2/(4 t)}\\ &= \frac1{2 \sqrt{\pi b t}} \int_0^{\infty} d\xi' \, e^{-\sqrt{b} \xi'} \left ( e^{-(\xi-\xi')^2/(4 t)} + e^{-(\xi+\xi')^2/(4 t)} \right ) \end{align}$$
There is some algebra in evaluating the last integral; I will leave this for the reader. The final result is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \, \, \frac1{p \sqrt{p+b}} e^{-\xi \sqrt{p+b}} e^{p t} = \\ \frac{e^{-\sqrt{b} \xi}}{\sqrt{b}} - \frac{e^{b t}}{\sqrt{b}} \frac12 \left [ e^{-\sqrt{b} \xi} \operatorname{erfc}{\left ( \sqrt{b t} - \frac{\xi}{2 \sqrt{t}} \right )} +  e^{\sqrt{b} \xi} \operatorname{erfc}{\left (  \sqrt{b t} + \frac{\xi}{2 \sqrt{t}}  \right )} \right ]  $$
where erfc denotes the complementary error function.
