This is a repost from Math Stack-exchange where I did not manage to get an answer. https://math.stackexchange.com/questions/1491027/inverse-laplace-transform-of-a-hypergeometric-function

I managed to solve an initial value problem in the Laplace domain in terms of a special function

$ F(s) = c_2 \frac{1}{{{\left( {{s}^{1 +\beta}}-1\right) }^{\frac{1}{\beta+1}}}}+ c_1 \frac{s}{{{\left( {{s}^{1 +\beta}}-1\right) }^{\frac{1}{\beta+1}}}} \ {_{2}{F}_{1}}\left( \frac{1}{\beta+1},\frac{\beta}{\beta+1}; \frac{1}{\beta+1}+1;{{s}^{\beta+1}}\right) $ where $ 0 \leq \beta \leq 1$ and $ _{2}{F}_{1}$ if the hypergeometric function.

However, I am unable to find the ILT or give an approximation in the time domain in the general case.

Is there a way to invert the equation or at least to give an approximation for short times?

  • $\begingroup$ Is the inverce Laplace transform knmown in this case? I wanted to glue the solution in the real domain to another one valid in a restricted interval. $\endgroup$ – Dimiter P Feb 6 '16 at 17:56

Using known reference books we derive that in this case $_2F_1(a,1-a;1+a;z)$ is reduced to incomplete Beta-function $B_{\frac{1-z}{2}}(a,a)$. That is much simpler and easy to estimate exactly or numerically.

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