# What is the series expansion of the $n$-th derivative of this : $\frac{d^n}{dx^n}\int{(e^{-x²})}^{\text{erf}(x)}dx$

$$\newcommand{\erf}{\operatorname{erf}}$$ The computation of $$\frac{d^n}{dx^n}\int{(e^{-x²})}^{\erf(x)}dx$$ with wolfram alpha we have for $$n=1, n=2, ..n=4$$ interesting expansion which seems present altern series such that exploit the formula for the constant $$\pi$$ , Now My attempt fails to get the series expansion of that integral at $$x=0$$ since I don't know any thing about it's closed form. Now my question here is :

Question :What is the series expansion of the $$n$$-th derivative of that function(integral) ? and what is the convergence rate (how many digits of $$\pi$$ can be computed at each increment in summation)?

• $n=1,2,3,4,...$ ? Why dont you write $$\frac{{\rm d}^{n-1}}{{\rm d}x^{n-1}} \left(e^{-x^2}\right) ^{{\erf}(x)} \, ?$$ – Diger Aug 18 '18 at 11:05
• Thanks , yes it is the same – zeraoulia rafik Aug 18 '18 at 11:06

$$\newcommand{\erf}{\operatorname{erf}}$$I don't know what you expect from the series expansion, but maybe for the start you can just use the series for each functions and write $$e^{-x^2 \erf(x)} = \sum_{m=0}^\infty x^m \substack{\sum_{k_1=0}^\infty \cdots \sum_{k_n=0}^\infty \\ \frac{m-2(k_1 + ... + k_n)}{3} \stackrel{!}{=} \, n \in {\mathbb N}} \frac{(-1)^{n+k_1+...+k_n}\left(\frac{2}{\sqrt{\pi}}\right)^n}{n! k_1!\cdots k_n! (2k_1+1)\cdots (2k_n+1)} \, .$$ The sums over $$k_1 \cdots k_n$$ are in fact finite sums, because $$n\geq 0$$ limited by $$k_{\rm max}=\left\lfloor \frac{m}{2} \right \rfloor$$.

Notation: The inner sum means, that the summation over $$k_1 \cdots k_n$$ is constrained by the requirement $$\frac{m-2(k_1 + ... + k_n)}{3} = \frac{m-2k}{3} = \, n \geq 0 \in {\mathbb N} \, .$$

For $$m=0$$ all the $$k$$ have to be zero in order for $$n=0$$ and this gives the trivial term $$1$$.

For $$m=1$$, $$k$$ values $$>0$$ lead to negative $$n$$ and $$k=0$$ leads to $$n=\frac{1}{3} \notin {\mathbb N}$$.

For $$m=2$$, $$k=0$$ or $$k=1$$. $$k=0$$ leads to a contradiction and $$k=1$$ leads to $$n=0$$ which is the empty sum (which starts at $$k_1$$ and not $$k_0$$).

$$m=3$$ is the first non-trivial case. $$k$$ must be either $$0$$ or $$1$$ for $$n\geq 0$$. $$k=1$$ again leads to $$n=\frac{1}{3}$$ so we only have $$k=0$$ in which case $$n=1$$ and the sum is actually just $$\sum_{\substack{k_1=0 \\ k_1 \stackrel{!}{=}0}}^1 \frac{(-1)^{1+k_1}\frac{2}{\sqrt{\pi}}}{1!k_1!(2k_1+1)} = -\frac{2}{\sqrt{\pi}} \, .$$

$$m=4$$ is again trivial, because $$k=0,1,2$$ leads to $$n=\frac{4}{3},\frac{2}{3},0$$.

The sums start to get more involved at $$m=9$$. Now $$k=0,1,2,3,4$$ leads to $$n=3,\frac{7}{3},\frac{5}{3},1,\frac{1}{3}$$ so there are 2 valid $$n$$ \begin{align} n=1: &\sum_{\substack{k_1=0 \\ k_1\stackrel{!}{=}3}}^{4} \frac{(-1)^{1+k_1}\frac{2}{\sqrt{\pi}}}{1!k_1!(2k_1+1)} = \frac{2}{\sqrt{\pi}} \frac{1}{3!(2\cdot3+1)} = \frac{1}{21 \sqrt{\pi}} \\ n=3: &\substack{\sum_{k_1=0}^4 \sum_{k_2=0}^4 \sum_{k_3=0}^4 \\ (k_1 + k_2 + k_3) \stackrel{!}{=} \, 0} \frac{(-1)^{3+k_1+k_2+k_3}\left(\frac{2}{\sqrt{\pi}}\right)^3}{3! k_1! k_2! k_3! (2k_1+1) (2k_2+1) (2k_3+1)} \\ &=-\frac{4}{3\pi^{3/2}} \, . \end{align}

There is a pattern in the $$n$$: For odd $$m$$ at $$m=6i-3$$ the number of allowed $$n$$ increases by $$1$$ of which there are $$i$$ in total. For instance at $$m=9$$ we had $$n=1,3$$. At $$m=15$$ we have $$n=1,3,5$$ and so on.

Similarly for even $$m$$ at $$m=6i$$ the number of allowed $$n$$ equals $$i$$ whose values take $$n=2,4,6,8,\dots$$.

• That is some pretty funky summation notation, and could probably use some explaining. – Theoretical Economist Aug 18 '18 at 12:08
• @Diger , Thanks , could show more how you got that , I want some steps explanation – zeraoulia rafik Aug 18 '18 at 12:53
• added some notation infos. – Diger Aug 18 '18 at 14:21
• Thanks , it's stay to know about rate of convergence as stated above in my question – zeraoulia rafik Aug 18 '18 at 14:28
• radius of convergence should be $\infty$. – Diger Aug 18 '18 at 15:49