Inverse Laplace Transform

  • Thread starter iiternal
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  • #1
iiternal
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Hi, all.
I am doing an inverse Laplace transform and meeting some difficulty.
InverseLaplaceTransform[1/(Sqrt+a)].
Using Mathematica I found the result, however, I failed to prove it. I know it should be like
\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{1}{\sqrt{s}+a}ds
and I can take \gamma =0.
But, then, I cannot continue. The sqrt on the denominator killed me.
Thank you very much.
 

Answers and Replies

  • #2
timthereaper
479
33
From what I can tell, there is no good antiderivative of e^(st)/(Sqrt+a). If you can find one, I'd love to know.
 
  • #3
iiternal
13
0
From what I can tell, there is no good antiderivative of e^(st)/(Sqrt+a). If you can find one, I'd love to know.


Well, by using Mathematica, I got a result expressed in terms of error functions. So I guess I have to transform it into error functions. But, how?
 
  • #4
timthereaper
479
33
Well, e^(st) can possibly be expressed as e^((Sqrt^2)*t), which when integrated with respect to s is similar to the error function. That might be the way to attack this.
 

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