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I have a Laplace transformed function that I'd like to transform back. It's quite a complex function however, which is why I am stuck:

$$C(x,s) = \left(\frac{m}{s}\right)\exp\left(\lambda^2\frac{\left(v-\sqrt{\frac{4 D s}{\lambda^2} - v^2}\right)}{2 D} x\right)$$

I know the inverse Laplace transform of a less complex, but comparable, function:

$$C(x,s) = \left(\frac{m}{s}\right) \exp\left(\frac{s}{k} x\right)$$

The solution of this would be (derived from results described in the more extensive Laplace transform tables):

$$C(x,t) = m \left(1 - \text{erf}\left(\frac{x}{\sqrt{4 k t}}\right)\right)$$

However, with the more difficult function it won't be that easy. I was trying to rewrite (split and simplify) the power term in order to make it simpler, but it doesn't really work.

I am new to the field of Laplace Transforms, so I am quite blind to the directions I could go. Does any of you have an idea on where to start (e.g., which method to use) to solve it?

I'd prefer to solve it by hand, but I am also interested on how to solve this with, e.g., Matlab. I tried to following piece of code, but it just returns my input:

syms c s k real
F = (c/s)*exp(-x*(sqrt(s/k)))
ilaplace(F,s)
horchler
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Phoneheads
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    See this: http://math.stackexchange.com/questions/350742/find-the-greens-function-and-solution-of-a-heat-equation-on-the-half-line/350758#350758 – Ron Gordon Dec 15 '15 at 14:19
  • Thank you for the link. I wish I could say it was helpful, but to be honest it confuses me more. I have little mathematical background and know nothing yet about contour integration. Is this something that is highly recommended to look into or are there approaches that are better understandable for a layman/maths newbie like me? It would be great for example if something would be possible with rewriting the formula. – Phoneheads Dec 15 '15 at 14:56
  • If you want to derive the expression for the ILT of the expression you have you need to perform some integration in the complex plane. Contour integration techniques are standard and way easier than anything I've ever seen. If you are not equipped then perhaps tables and the convolution theorem might help. My link has an example of this at the end. – Ron Gordon Dec 15 '15 at 14:59
  • Thank you, I will have to take a look into that then. – Phoneheads Dec 15 '15 at 15:46

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