Taylor expansion of a gaussian integral I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}
So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.
My questions are:
How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)
Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?

This is not a gaussian integral.
$$f(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} e^{-\frac{xy^4}{24}-\frac{y^2}{2}} \, dy=\frac{\sqrt{3}}{\sqrt{2\pi}}\frac{e^{\frac{3}{4 x}} }{\sqrt{x}}K_{\frac{1}{4}}\left(\frac{3}{4 x}\right)$$
Using the asymptotics of the Bessel function (since $x \to 0 \implies \frac 1 x \to \infty$), you should have
$$f(x)=1-\frac{x}{8}+\frac{35 x^2}{384}-\frac{385 x^3}{3072}+\frac{25025
   x^4}{98304}-\frac{1616615 x^5}{2359296}+O\left(x^6\right)  $$
If you make $x=\frac 1{10}$, this truncated series will give
$$f\left(\frac{1}{10}\right)=\frac{46634068277}{47185920000}=0.9883047$$ while the exact value is
$$f\left(\frac{1}{10}\right)=e^{15/2} \sqrt{\frac{15}{\pi }} K_{\frac{1}{4}}\left(\frac{15}{2}\right)=0.9883064$$
If you want more terms, let $x=\frac{3}{4 z}$ which makes
$$f(z)=\sqrt{\frac{2}{\pi }} e^z \sqrt{z} K_{\frac{1}{4}}(z)$$ with $z \to \infty$. Have a look at this paper (equation $(1.10)$) to have
$$f(z)=\Bigg[\sqrt{\frac{2}{\pi }} e^z \sqrt{z}\Bigg]\sqrt{\frac{\pi }{2z}} e^{-z} \sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}=\sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}$$
From equation $(1.9)$
$$ a_n\left({\frac 14}\right)=(-1)^n\frac{  \cos \left(\frac{\pi }{4}\right) \Gamma
   \left(n+\frac{1}{4}\right) \Gamma \left(n+\frac{3}{4}\right)}{\pi\, 2^n\, \Gamma (n+1)}$$
Make $z=\frac 3{4x}$ to recover the first expansion.
Edit
Without using the exact solution (as I did), you could have done the same work expanding first the integrand as a Taylor series around $x=0$
$$e^{-\frac{xy^4}{24}-\frac{y^2}{2}}=\sum_{n=0}^\infty\frac {(-1)^n}{24^n\,n!} e^{-\frac{y^2}{2}} y^{4 n} x^n$$ and use that
$$\int_{-\infty}^\infty e^{-\frac{y^2}{2}} y^{4 n}\,dy=2^{2 n+\frac{1}{2}} \Gamma \left(2 n+\frac{1}{2}\right)$$ This makes
$$f(x)= \frac{1}{\sqrt{\pi}}\sum_{n=0}^\infty\frac {(-1)^n}{6^n\,n!}  \Gamma \left(2 n+\frac{1}{2}\right) x^n$$ ad the same result.
