Does this Nelson formula for Feynman integral have mistakes? In this paper (Maroun's PhD dissertation, 2013) at page 46 the following formula is given (apparently without a reference):
$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{\frac{1}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$
Now I am trying to verify the formula. If I take $a=b=i/2$, $s=p=1$ the left hand side becomes $1$ while right hand side becomes $8/3$. 
It had been suggested on MathOverflow, that the formula has a mistake, and the right-hand part should contain $ba^{-p/s}$ instead of $ba^{1/s}$. In this case with the above-mentioned data the equality holds, but still it does not work for $a=-b$; $p=s$. 
Is there some error in the formula? Is this formula even well-known so to look somewhere for the correct form?
 A: It is correct that there is a typo in the right hand side series expression. The term in the numerator of the series inside of parenthesis raised to the n power should be
$$
\left(iba^{-\frac{p}{s}}\right)^n.
$$
A: Taylor series Expansion of $e^{ibx^p}$ yields
$\sum_{n=0}^\infty \frac{(ib)^n}{n!}x^{np}$. 
Then you have to evaluate every term of the series by a Substitution. You can use the Substitution $u = -iax^s$ which yields $x = (i\frac{u}{a})^{1/s}$ and 
$dx = e^{\frac{i \pi}{2s}}a^{-1/s}u^{1/s-1}/s$.
The integral $I_n = \int_0^\infty dx x^{np}e^{iax^s}$ turns to $I_n =  \int_0^\infty \frac{du}{su} (i\frac{u}{a})^{np/s+1/s} e^{-u} = \int_0^\infty \frac{du}{su}(ia^{-1})^{np/s+1/s}u^{np/s+1/s} e^{-u} = e^{i \pi \frac{np+1}{2s}}a^{-np/s}/(a^{1/s}s) \int_0^\infty du u^{np/s+1/s-1} e^{-u} = \frac{e^{i \pi \frac{np+1}{2s}}{ \Gamma(\frac{np+1}{s})}a^{-np/s}}{a^{1/s}s}$.
Conclusion: You see that this series Expansion will take constants $(iba^{-p/s})^n$ that belong to the term $I_n$; the integral in your question has a small mistake.
The correct formula should be:
$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{-\frac{p}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$
