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Calculate the inverse Laplace transform of: $$\frac{1}{s\cdot(\sqrt{s}+1)} \cdot e^{-\sqrt{s} \cdot x}$$

My attempt at the solution was to break down the fraction with partial fraction decomposition as follows: $$\frac{1}{s \cdot (\sqrt{s}+1)} = \frac1s-\frac{1}{\sqrt{s}}+\frac{1}{1+\sqrt{s}}$$ Then the first part can be easily computed from the table or by using some software: $$\mathcal{L}^{-1}( s^{-1} \cdot e^{-\sqrt{s} \cdot x} )= 1-erf(\frac{x}{2\sqrt{t}})$$ However the second part is not at all trivial, as I was unable to find any coherent answer to the problem: $$\mathcal{L}^{-1}( (-\frac{1}{\sqrt{s}}+\frac{1}{1+\sqrt{s}}) \cdot e^{-\sqrt{s} \cdot x} )$$ Could someone, please, guide towards the correct answer? I have tried using computer algebra systems such Mathematica, but nothing seems to work.

***Treat x as a constant with x>0.

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  • $\begingroup$ in the first equation, in the exponential what is $x$ in $\sqrt{s}x$? $\endgroup$ – Picaud Vincent Oct 2 '18 at 20:54
  • $\begingroup$ I am sorry I have not mentioned that, x is an arbitrary constant such that x>0. $\endgroup$ – statwanderer Oct 2 '18 at 21:31
  • $\begingroup$ Using Mathematica I get $\left(t+\frac{x^2}{2}\right) \text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)-\frac{\sqrt{t} x e^{-\frac{x^2}{4 t}}}{\sqrt{\pi }}$ (I have to leave right now, sorry. I hope it can help) $\endgroup$ – Picaud Vincent Oct 2 '18 at 21:40
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    $\begingroup$ Do you know what the "Convolution Theorem" is?(also use \cdot instead of * for product because * if for convolution) $\endgroup$ – ℋolo Oct 2 '18 at 21:57
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    $\begingroup$ @PicaudVincent. It seems you are wrong ?.Check: FullSimplify[ LaplaceTransform[(t + x^2/2)*Erfc[x/(2 Sqrt[t])] - (Sqrt[t]*x)/Sqrt[ Pi]*Exp[-x^2/(4 t)] // Expand, t, s], Assumptions -> {x > 0, s > 0}] . $\endgroup$ – Mariusz Iwaniuk Oct 2 '18 at 22:15
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$$\color{red}{\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{\left(1+\sqrt{s}\right) s}\right](t)}=\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{1+\sqrt{s}}+\frac{e^{-\sqrt{s} x}}{s}-\frac{e^{-\sqrt{s} x}}{\sqrt{s}}\right](t)=\color{red}{\text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)-e^{t+x} \text{erfc}\left(\frac{2 t+x}{2 \sqrt{t}}\right)}$$ With CAS help:

$$\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{s}-\frac{e^{-\sqrt{s} x}}{\sqrt{s}}\right](t)=-\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{\pi } \sqrt{t}}+\text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)$$ It's hard to find $\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{1+\sqrt{s}}\right](t)$?

$$\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{1+\sqrt{s}}\right](t)=\mathcal{L}_s^{-1}\left[\mathcal{L}_a^{-1}\left[\frac{e^{-\sqrt{s} x}}{a+\sqrt{s}}\right](q)\right](t)=\mathcal{L}_q\left[\mathcal{L}_s^{-1}\left[e^{-q \sqrt{s}-\sqrt{s} x}\right](t)\right](1)=\mathcal{L}_q\left[\frac{e^{-\frac{(q+x)^2}{4 t}} (q+x)}{2 \sqrt{\pi } t^{3/2}}\right](1)=\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{\pi } \sqrt{t}}-e^{t+x} \text{erfc}\left(\frac{2 t+x}{2 \sqrt{t}}\right)$$

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  • $\begingroup$ If I may ask, out of curiosity, how did you come up with such ingenious solution to the second Laplace transform? $\endgroup$ – statwanderer Oct 3 '18 at 12:50
  • $\begingroup$ @RenatSergazinov. This trick is not mine. A user of math.stackexchange.com used it, but I do not remember which one. $\endgroup$ – Mariusz Iwaniuk Oct 3 '18 at 13:56

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