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I need to get the Inverse Laplace Transform badly for the following function $$\frac{1}{\beta + \sqrt{p}} e^{-\alpha \sqrt{p + \gamma}},$$ $\alpha,\, \beta,\, \gamma$ being some parameters.

I have looked through Erdelyi's book of tables and found only the expression for $$\mathcal{L}^{-1}_{p} \left( \frac{1}{\beta + \sqrt{p}} e^{-\alpha \sqrt{p }} \right)(t) = \frac{1}{\sqrt{\pi t}}e^{-\frac{\alpha^{2}}{4t}} - \beta e^{\alpha \beta + \beta^{2} t} \mathbb{Erfc} \left( \frac{\alpha}{2\sqrt{t}} + \beta \sqrt{t} \right).$$

My idea was to understand how to compute the known formula from Erdelyi and then to deduce the one I need. However, I didn't even succeed with understanding the derivation of the known formula.

Here is what I have already tried:

  • Bromwich integral;
  • Convolution: $\mathcal{L}^{-1}_{p} \left( \frac{1}{\beta + \sqrt{p}} e^{-\alpha \sqrt{p }} \right) (t) =\mathcal{L}^{-1}_{p} \left( \frac{1}{\beta + \sqrt{p}} \right) * \mathcal{L}^{-1}_{p} \left( e^{-\alpha \sqrt{p}} \right) (t)$;
  • Applying the formula for a square-root: $\mathcal{L}^{-1}_{p} \left( F(\sqrt{p}) \right) (t) = \frac{1}{2\sqrt{\pi} t^{3/2}} \int_{0}^{\infty}x e^{-x^{2}/4t} f(x) dx$, where $F(p) = \frac{1}{\beta + p} e^{-\alpha p}$ and $\mathcal{L}^{-1}_{p} \left( F(p) \right) (t) = f(t) = e^{-\beta (t-\alpha)} I_{ \{ t > \alpha \} }$. But this method is not very useful for performing the initial task because of the absence of the possibility to make a shift in one of two square-roots (as I am aware, only both can be performed simultaneously).

In the two first methods, I cannot guarantee that I didn't miss something important and therefore cannot obtain the explicit formulae.

To sum the things up, I would appreciate the step by step derivation for both the initial task and for the easier one being the second task.

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Using the convolution theorem:

$$\text{f}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\exp\left(-\alpha\cdot\sqrt{\gamma+\text{s}}\right)}{\beta+\sqrt{\text{s}}}\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\beta+\sqrt{\text{s}}}\right]_{\left(t\right)}\space*\space\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\alpha\cdot\sqrt{\gamma+\text{s}}\right)\right]_{\left(t\right)}$$

And we get that:

$$\text{f}\left(t\right)=\left(\frac{1}{\sqrt{\pi}{\sqrt{t}}}-\beta e^{\beta^2 t}\cdot\text{Erfc}\left(\beta\sqrt{t}\right)\right)\space*\space\left(\frac{\alpha e^{-\frac{\alpha^2}{4t}-t\gamma}}{2\sqrt{\pi}t^\frac{3}{2}}\right)$$

So, when we undo the convolution we need to compute:

$$\text{f}\left(t\right)=\int_0^t\left(\frac{1}{\sqrt{\pi}{\sqrt{\tau}}}-\beta e^{\beta^2 \tau}\cdot\text{Erfc}\left(\beta\sqrt{\tau}\right)\right)\left(\frac{\alpha e^{-\frac{\alpha^2}{4\left(t-\tau\right)}-\left(t-\tau\right)\gamma}}{2\sqrt{\pi}\left(t-\tau\right)^\frac{3}{2}}\right)\space\text{d}\tau$$

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  • $\begingroup$ Thank you for the answer but that is utterly clear. The question is how to perform the integration explicitly in the case. I have no clue. Moreover, I do not understand how to carry out the integration even in the simpler case: $$f(t) = \int_{0}^{t} \left( \frac{1}{\sqrt{\pi \tau}} - \beta e^{\beta^2 \tau} \mathbb{Erfc(\beta \sqrt{\tau})} \right) \left( \frac{\alpha e^{-\frac{\alpha^2}{4(t-\tau)}}}{2\sqrt{\pi} (t-\tau)^{3/2}} \right) d\tau$$ @Jan Eerland $\endgroup$ – Dmitrii Feb 20 '17 at 17:36

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