behavior of :$\int_{-1}^1x^{2k} \operatorname{erf}(x)^k \,dx $?

Question

I'm interesting for the evaluation of $\int_{-1}^1 x^{2k} \operatorname{erf}(x)^k \,dx=0$ with $l$ is a real number , I want to get closed form of that integral , i'm coming up to define it as :$\int_{-1}^1 x^{2k} \operatorname{erf}(x)^k \,dx=0$ if $k$ is odd integer number and else ( equal to $\beta \neq 0$ ) if $k$ is even number ,Now i have two question for asking :

Question (1):

Is really :$\int_{-1}^{1}x^{2k} \operatorname{erf}(x)^k \, dx=0$ if $k$ is odd integer number and else ( equal to $\beta \neq 0$ ) if $k$ is even number ?

Question (2): Can i get a closed form of it in the case of $k$ is even integer over $[-1,1]$ ?

Edit 01: I have edit the question (2) without changing the meaning according to the given answer such that i want it's closed form in the range $[-1,1]$ if it is possible

Edit 02 I have took Boundaries since it's complicated to get closed form with determinate boundaries

Answer 1

$$I_k=\int_{-1}^1 x^{2k}\, \left[\text{erf}(x)\right]^k \,dx$$ is $0$ if $k$ is odd. So, we need to focus on $$I_{2k}=\int_{-1}^1 x^{4k}\, \left[\text{erf}(x)\right]^{2k} \,dx$$ which could be approximated, as I wrote in an answer to your previous question, using $$\left[\text{erf}(x)\right]^2\approx 1-e^{-a x^2}\qquad \text{with}\qquad a=(1+\pi )^{2/3} \log ^2(2)$$ making $$I_{2k}=\int_{-1}^1 x^{4k}\, \left(1-e^{-a x^2}\right)^{k} \,dx$$ to be developed using the binomial expansion.

So, in practice, we face the problem of $$J_{n,k}=\int_{-1}^1 x^{4k}\,e^{-n a x^2}\,dx$$ The antiderivative $$\int x^{4k}\,e^{-n a x^2}\,dx=-\frac{1}{2} x^{4 k+1} E_{\frac{1}{2}-2 k}\left(a n x^2\right)$$ where appears the exponential integral function. Using the bounds, this reduces to $$J_{n,k}=-E_{\frac{1}{2}-2 k}(a n)$$ and leads to "reasonable" approximation as shown in the table below $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.22870436048 & 0.22959937502 \\ 2 & 0.08960938943 & 0.08997882179 \\ 3 & 0.04400808083 & 0.04418398568 \\ 4 & 0.02389675159 & 0.02398719298 \\ 5 & 0.01374034121 & 0.01378897319 \\ 6 & 0.00819869354 & 0.00822557475 \\ 7 & 0.00502074798 & 0.00503586007 \\ 8 & 0.00313428854 & 0.00314286515 \\ 9 & 0.00198581489 & 0.00199069974 \\ 10& 0.00127304507 & 0.00127582211 \end{array} \right)$$

Edit

Another approximation could be obtained using the simplest Padé approximant of the error function $$\text{erf}(x)=\frac{2 x}{\sqrt{\pi } \left(1+\frac{x^2}{3}\right)}$$ which would lead to $$I_{2k}=\int_{-1}^1 x^{4k}\, \left[\text{erf}(x)\right]^{2k} \,dx$$ $$I_{2k} \approx\frac 2{6k+1}\,\left(\frac{4}{\pi }\right)^k\,\, _2F_1\left(2 k,\frac{6k+1}{2}; \frac{6k+3}{2};-\frac{1}{3}\right)$$ slightly less accurate than the previous one.

Continuing with Padé approximant $$\text{erf}(x)=\frac{\frac{2 x}{\sqrt{\pi }}-\frac{x^3}{15 \sqrt{\pi }}}{1+\frac{3 x^2}{10}}$$ we should get $$I_{2k}=\int_{-1}^1 x^{4k}\, \left[\text{erf}(x)\right]^{2k} \,dx$$ $$I_{2k}\approx\frac 2{6k+1}\,\left(\frac{4}{\pi }\right)^k\,\, F_1\left(\frac{6k+1}{2};-2 k,2 k; \frac{6k+3}{2};\frac{1}{30},-\frac{3}{10}\right)$$ where appears the the Appell hypergeometric function of two variables.

Answer 2

For $k=2$, Maple says the integral is $${\frac { \left( 16\, \left( {\rm erf} \left(l\right) \right) ^{2 }{l}^{5}{\pi}^{3/2}{{\rm e}^{2\,{l}^{2}}}+32\,{\rm erf} \left(l\right) {{\rm e}^{{l}^{2}}}\pi\,{l}^{4}-43\,\pi\,\sqrt {2}{{\rm e}^{2\,{l}^{2} }}{\rm erf} \left(\sqrt {2}l\right)+64\,{\rm erf} \left(l\right){ {\rm e}^{{l}^{2}}}\pi\,{l}^{2}+16\,\sqrt {\pi}{l}^{3}+64\,{\rm erf} \left(l\right){{\rm e}^{{l}^{2}}}\pi+44\,\sqrt {\pi}l \right) {{\rm e} ^{-2\,{l}^{2}}}}{40\;{\pi}^{3/2}}} $$ It doesn't find a closed form for $k=4$, nor does Wolfram Alpha, and I think it's quite unlikely that one exists.