Proving the Inverse Laplace Transform of 1/(sqrt(s)+a)

[iiternal]

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.

 

[timthereaper]

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.

 

[iiternal]

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?

 

[timthereaper]

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.