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 $x \geq1$ ?
 A: Indeed the radius of convergence of the function $f$ is zero, which means that the series
$$ \sum_{k=0}^{\infty} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k $$
diverges for all $z \neq 0$. Although this can be directly verified by using Stirling's approximation, a more fundamental reason for this is that $0$ is a branch point of $f(z)$. Consequently, there is no way $f(z)$ extends to an analytic function about $0$. For instance, the closed-form
$$ f(z) = \sqrt{\frac{3}{2\pi z}} \, e^{3/4z} K_{1/4}(3/4z) \tag{*} $$
in Claude Leibovici's answer has the branch cut $(-\infty, 0]$, as seen from the domain coloring plot of $\text{(*)}$ on the rectangle with the corners $\pm 5 \pm 5i$:

That said, your best hope for an asymptotic expansion of polynomial form is
$$ f(z) = \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k + \mathcal{O}(z^N) \qquad \text{as } z \to 0 \tag{1} $$
within Stolz angle for each fixed $N \geq 0$.

Addendum. Here is a proof that $\text{(1)}$ holds at least within the region $\operatorname{Re}(z) \geq 0$, although I suspect that this holds within larger sectors.
Note that for each fixed $N$, Taylor's theorem tells that for $\operatorname{Re}(z) \geq 0$,
\begin{align*}
&f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl[ \exp\left(-\frac{zy^4}{24}\right) - \sum_{k=0}^{N-1} \frac{1}{k!} \left( -\frac{zy^4}{24} \right)^k \Biggr] e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{z^N (1-t)^{N-1}}{(N-1)!} \left(-\frac{y^4}{24}\right)^N \exp\left(-\frac{zty^4}{24}\right) \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y,
\end{align*}
where the last line follows from the Lagrange form of the remainder. Taking absolute value,
\begin{align*}
&\left| f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \right| \\
&\leq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{\left| z \right|^N (1-t)^{N-1}}{(N-1)!} \frac{y^{4N}}{24^N} \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{(4N-1)!!}{N!24^N} \left|z\right|^N.
\end{align*}
This proves that the error term in $\text{(*)}$ is indeed $\mathcal{O}(z^N)$, at least within the region $\operatorname{Re}(z) \geq 0$.
A: 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.
A: Your function may be expressed in terms of the modified Bessel function of the second kind as follows:
$$
f(x)=\sqrt {\frac{3}{{2\pi x}}} e^{\frac{3}{{4x}}} K_{1/4} \left( {\frac{3}{{4x}}} \right).
$$
By the results of the paper http://dx.doi.org/10.1007/s10440-017-0099-0, for all $N\geq 0$ and $x$ with $|\arg x|<\frac{3\pi}{2}$, we have
$$
f(x) = \sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{(4n - 1)!!}}{{24^n n!}}x^n }  + R_N (x),
$$
where the remainder $R_N (x)$ satisfies
$$
\!\!\left| {R_N (x)} \right| \!\le\! \frac{{(4N - 1)!!}}{{24^N N!}}\left| x \right|^N 
\!\times\! \begin{cases} \!1 & \!\!\text{if } \; \left|\arg x\right| \leq \frac{\pi}{2}, \\ \!\min\! \left(\chi\!\left(N+\frac{1}{4}\right)+1,|\csc ( \arg x)|\right) & \!\!\text{if } \; \frac{\pi}{2} < \left|\arg x\right| \leq \pi, \\ \!\cfrac{\sqrt {2\pi \left( N + \frac{1}{4} \right)} }{\left| \cos (\arg x)\right|^{N + 1/4} } + \chi\!\left(N+\frac{1}{4}\right)+1 & \!\!\text{if } \; \pi < \left|\arg x\right| < \frac{3\pi}{2}. \end{cases}
$$
Here, for $p>0$,
$$
\chi (p) = \sqrt \pi  \frac{{\Gamma\! \left( {\frac{p}{2} + 1} \right)}}{{\Gamma \!\left( {\frac{p}{2} + \frac{1}{2}} \right)}},\quad \sqrt {\frac{\pi }{2}\left( {p + \frac{1}{2}} \right)}  \le \chi (p) \le \sqrt {\frac{\pi }{2}\left( {p + \frac{2}{\pi }} \right)} .
$$
As it was noted by others, the series cannot converge for any $x \neq 0$ because $x=0$ is a branch point of $f(x)$. However, it is an asymptotic expansion of $f(x)$ as $x\to 0$. As you can see, we have an explicit control over the error term. To obtain the best numerical approximation, stop the series at its numerically least term, i.e., at $N = \left[ {\frac{3}{{2\left| x \right|}}} \right]$ and estimate the error using the formulae above.
