2

I am trying to find the inverse Laplace transform of $e^\sqrt{as}$ for $a>0$. So we need to solve $\oint_B dz \: e^\sqrt{az} e^{z t}$ (Bromwich contour), but not sure how to start. How do we even see where the poles are?

Thank you

bart
  • 67
  • fyi, Mathematica says only sqrt[a]<0 InverseLaplaceTransform[Exp[Sqrt[a s]], s, t] is ConditionalExpression[-(Sqrt[a]/(E^(a/(4*t))*(2*Sqrt[Pi]*t^(3/2)))), Sqrt[a] < 0] – Nasser Nov 16 '13 at 07:45
  • See this solution, which illustrates that the problem is the branch point at $s=0$: http://math.stackexchange.com/questions/347933/compute-the-inverse-laplace-transform-of-e-sqrtz?rq=1 – Ron Gordon Nov 16 '13 at 07:57
  • What I am not sure about is on $C_5$ where does the minus sign go in front of $i\sqrt x$. – bart Nov 18 '13 at 06:44
  • @RonGordon Is that due to the branch cut? – bart Nov 18 '13 at 14:42
  • @bart: Yes, that's right. – Ron Gordon Nov 18 '13 at 14:55
  • @RonGordon Okay, but I am still not seeing why is $e^{-\sqrt{x e^{-i \pi}}}=e^{i\sqrt{x}}$ on $C_5$ – bart Nov 18 '13 at 15:06
  • @bart: $\sqrt{e^{i \pi}} = e^{i \pi/2} = i$. OTOH $\sqrt{e^{-i \pi}} = e^{-i \pi/2} = -i$ – Ron Gordon Nov 18 '13 at 15:11
  • @RonGordon Ah, thank you! – bart Nov 18 '13 at 16:36
  • @bart I suppose that there is a typo in the wording : probably it isn't $\exp(\sqrt{as})$, but $\exp(-\sqrt{as})$ Then, find the Laplace transform of $\exp\dfrac{-a/(4x)}{x^{3/2}}$ and see what is obtained.: – JJacquelin Jan 25 '16 at 14:55

0 Answers0