inverse laplace transform of $\frac{1}{s+b}e^{-x\sqrt{\frac{s}{k}}}$

Question

I am attempting to find the inverse laplace transform of $\frac{1}{s+b}e^{-x\sqrt{\frac{s}{k}}}$

The solution should be

$$\frac{e^{-bt}}{2} ( {e^{x\sqrt{\frac{-b}{k}}}\ erfc\left(\frac{x+2kt\sqrt{\frac{-b}{k}}}{2\sqrt{kt}}\right)+e^{-x\sqrt{\frac{-b}{k}}}erfc\left(\frac{x-2kt\sqrt{\frac{-b}{k}}}{2\sqrt{kt}}\right)})$$

I attempted changing $s+b$ to $s'$ which allowed me to find the inverse Laplace of $\frac{1}{s'}e^{-x\sqrt{\frac{s'-b}{k}}}$, however the solution to this was missing the $\frac{e^{-bt}}{2}$. I.e , the solution was $$( {e^{x\sqrt{\frac{-b}{k}}}\ erfc\left(\frac{x+2kt\sqrt{\frac{-b}{k}}}{2\sqrt{kt}}\right)+e^{-x\sqrt{\frac{-b}{k}}}erfc\left(\frac{x-2kt\sqrt{\frac{-b}{k}}}{2\sqrt{kt}}\right)})$$

I also tried using the convolution theorem with $F(s) = \frac{1}{b+s}$ and $G(s) = e^{-x\sqrt{\frac{s}{k}}}$, but was unable to solve the convolution integral.

Any help would be appreciated.

Answer 1

I'm very interested in the correct solution of this problem.
So I also tried the convolution approach

$$f(x,t)=\mathcal{L}_s^{-1}\left[\exp \left(-x \sqrt{\frac{s}{k}}\right)\right](t)=\frac{k x e^{-\frac{x^2}{4 k t}}}{2 \sqrt{\pi } \sqrt{k^3 t^3}}$$

$$g(t)=\mathcal{L}_s^{-1}\left[\frac{1}{s+b}\right](t)=e^{-b t}=\sum_{j=0}^{\infty} \frac{(-b t)^j}{j!}$$

Now we evaluate the convolution integral

$$\int_{0}^{t} f(x,\tau)\cdot g(t-\tau)\ d\tau=\int_0^{t} \sum_{j=0}^{\infty} \frac{k x e^{-\frac{x^2}{4 k \tau }} (-b (t-\tau ))^j}{2 \sqrt{\pi } j! \sqrt{k^3 \tau ^3}}\ d\tau$$

We exchange integration and infinite sum and get (Mathematica helps)

$$u(x,t)=\sum_{j=0}^{\infty} \int_0^{t} \frac{k x e^{-\frac{x^2}{4 k \tau }} (-b (t-\tau ))^j}{2 \sqrt{\pi } j! \sqrt{k^3 \tau ^3}}\ d\tau=\sum_{j=0}^{\infty} (-b t)^j \left(\frac{\, _1F_1\left(-j;\frac{1}{2};-\frac{x^2}{4 k t}\right)}{\Gamma (j+1)}-\frac{x\cdot \, _1F_1\left(\frac{1}{2}-j;\frac{3}{2};-\frac{x^2}{4 k t}\right)}{\Gamma \left(j+\frac{1}{2}\right) \sqrt{k t}}\right)$$

Does anyone know how to get rid of the infinite sum?

Visualization with $[b=1, k=1, 0\le j\le 18]$

HINT: Max. approximation error with $19$ summands is $\approx 3\cdot 10^{-12}$.