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 $x \geq1$ ?

  • 1
    $\begingroup$ I'd like to add something less 'technical' to the good answers below. Your integral diverges for all $x<0$. The behavior of the integral changes fundamentally between $x<0$ and $x>0$. Your expansion is about the singular point $x=0$, and so (in general) will result in a divergent series, or a series in non-integer powers of $x$ (which you have excluded by Taylor expanding). So we could anticipate the divergence of the series in advance. It can be shown that this series is asymptotic. If such things interest you, look up the book: "Advanced mathematical methods" by Carl Bender. $\endgroup$
    – Sal
    Commented Feb 28, 2021 at 14:46

3 Answers 3


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 $\pm 5 \pm 5i$:

Domain coloring plot of f

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 $\operatorname{Re}(z) \geq 0$.

  • $\begingroup$ Thank you for your answer ! However, if the error term in (1) is $O(z^N)$, then why doesn't this error converge to zero ? (because $z << 1$). $\endgroup$ Commented Feb 28, 2021 at 15:27
  • 1
    $\begingroup$ @Einsteinwasmyfather To show that the series is convergent, you take the limit $N\to +\infty$ and not $x\to 0$. However, the bound for the error term tends to infinity as $N\to + \infty$ (provided $x \neq 0$), so it does not give you convergence. For an asymptotic series the requirement is that the $N$th error term is of order $z^N$ for each fixed $N$ as $z\to 0$. For a convergent series you consider the case that $z$ is fixed and $N\to +\infty$. $\endgroup$
    – Gary
    Commented Feb 28, 2021 at 18:54

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.


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.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .