This question is related to my question here such that i want to find a closed form of $\int_{-1}^1 x^{2k} (\operatorname{erf}(x))^k \,dx $ , for $k$ is even integer because for odd integer is $0$ as we have integrand of odd function , Really for $k=2$ it gives a nice closed form as shown here in wolfram alpha, Now my question here is :

Is it possible to get a Closed form of :$\int_{-1}^1 x^{2k} \operatorname{erf}(x)^k \,dx $ for $ k$ is even integer?

Note: The Motivation of this question is to find a Series representation of :$\int _{0}^{t}\exp(-x^2 \operatorname{erf}(x))dx$

  • 2
    $\begingroup$ As a general rule of thumb: since $1$ and $-1$ are not spacial points for the integrand, his definite integral is likely just as difficult as the indefinite integral. So: $\int x^4 \text{erf}(x)^2 dx$ has a known closed form, plug in and get your definite integral. But for $\int x^8 \text{erf}(x)^4 dx$ you are probably out of luck. $\endgroup$ – Gerald Edgar May 31 '18 at 23:57
  • $\begingroup$ What for? Isn't it art for art's sake? $\endgroup$ – user64494 Jun 1 '18 at 9:03
  • $\begingroup$ No, it is not art for art's sake. $\endgroup$ – Liviu Nicolaescu Jun 1 '18 at 13:15
  • $\begingroup$ @Liviu Nikolaescu: Please, base your statement. $\endgroup$ – user64494 Jun 1 '18 at 17:50
  • $\begingroup$ @user64494, Probably Liviu Nicolaescu think that is easy for evaluation $\endgroup$ – zeraoulia rafik Jun 1 '18 at 17:52

I understand from the OP that the motivation for this question is to find a series expansion in powers of $t$ of $$I(t)=\int _{0}^{t}\exp(-x^2 \operatorname{erf}(x))dx=\sum_{p=1}^\infty c_p t^p.$$ The coefficients $c_p=p^{-1}d_{p-1}$ follow from the series expansion $e^{-x^2\,{\rm erf}\,x}=\sum_{p=0}^\infty d_p x^p$, resulting in $$I(t)=\sum_{p=1}^\infty c_p t^p=t-\frac{t^4}{2 \sqrt{\pi }}+\frac{t^6}{9 \sqrt{\pi }}+\frac{2 t^7}{7 \pi }-\frac{t^8}{40 \sqrt{\pi }}-\frac{4 t^9}{27 \pi }+\frac{(\pi -28) t^{10}}{210 \pi ^{3/2}}+O\left(t^{11}\right).$$ The series $I(1)=\sum_{p=1}^\infty c_p$ seems to converge:

Plot of $I_N=\sum_{p=0}^{N} c_p $ as a function of $N$ up to $N=25$. The value of $I_{25}=0.8162$ agrees with $I(1)=0.816377$ to three decimal places. For $N=50$ the agreement is up to six decimal places.

  • $\begingroup$ But is it possible to get it for n th term and it's seems that is a formel power series ? $\endgroup$ – zeraoulia rafik Jun 1 '18 at 13:05
  • $\begingroup$ The value of I(50) from wolfram alpha is : 0.972107 , wolframalpha.com/input/… $\endgroup$ – zeraoulia rafik Jun 1 '18 at 17:02
  • $\begingroup$ @zeraouliarafik -- when I say $I_{50}=0.816377$ I am referring to the sum $I_{N}=\sum_{p=0}^N c_p$, which is the series expansion of $I(t)$ for $t=1$ to order 50; what you have evaluated with Wolfram alpha is $I(t)$ for $t=50$, which is a different thing. $\endgroup$ – Carlo Beenakker Jun 1 '18 at 18:08
  • $\begingroup$ I'm confused how you calculated the coeffecients of series using error function ? $\endgroup$ – zeraoulia rafik Jun 2 '18 at 2:50

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