Borel summation

[JmsNxn]

\[
**\begin{align}
\int_0^\infty e^{-x} F(\sqrt{t}x)\,dx &= \sum_{k=0}^\infty b_k t^k\\
&= t^{-1/2}\int_0^\infty e^{-ut^{-1/2}} F(u)\,du
\end{align}
\]

That's a relief, James, this formula finally works very well numerically!
This is the function under the integral t=2:

And at a first glance also gives very good results for the half-iterate.

YES!

I plan to do a write up soon; I'll add some more flavour to that formula. We've opened the mellin transform/fourier transform/laplace transform flood gates!!!

Also, note that this form of the answer is much better for \(t \approx 0\), and in the attracting petal. We want \(e^{-ut^{-1/2}}\) to be as small as possible--and \(t\) to be within the region of the asymptotic expansion. To make this work "globally" is trickier, because we need to analytically continue the laplace transform--which sounds harder than it is