Does $\sum_{n=-\infty }^\infty n^2 \exp(-n^2 \operatorname{erf}(n^2))=1$?

Question

I have run this sum in mathematica , it returned me this form which related to Elliptic theta function as:

$$ e^{-\operatorname{erf}(n^2)} \operatorname{EllipticTheta}^{(0,0,1)}(3,0,e^{-\operatorname{erf}(n^2)}),$$ With Elliptic theta I think present [Jacobi theta function], The below plot present it partial sum in the range $(-10,10)$ seems like a CDF for The following Function(PDF) such that it can be normalized easly :$F(x)=x^2 \exp(-x^2 \operatorname{erf}(x^2))$ such that integrand of that PDF over $(-\infty,\infty)$ gives this result :$0.989536$ , Now my question here is : Does Really :$$ \int_{-\infty}^\infty x^2 \exp (-x^2 \operatorname{erf}(x^2)) \, dx =\sum_{n=-\infty}^\infty n^2 \exp(-n^2 \operatorname{erf}(n^2)) \text{?}$$

Note: For My attempt I have approximated Erf using Pade Approximant using inside of that the Theta Jacobie function but I didn't succeed to get the above closed Form , Any Way ?

Answer 1

Using Mathematica It seems that $$\sum _{n=-\infty }^{\infty } n^2 \exp \left(-n^2 \mathrm{erf}\left(n^2\right)\right)=1.00984.$$ Here the plot of partial sum with some corresponding values: However, $$\int_{-\infty }^{\infty } x^2 \exp \left(-x^2 \mathrm{erf}\left(x^2\right)\right) \, dx =0.989535665531394\ldots$$

Answer 2

In fact, $$ \sum_{n=-\infty}^\infty n^2 \exp(-n^2 \operatorname{erf}(n^2)) \ge \sum_{n=-2}^2 n^2 \exp(-n^2 \operatorname{erf}(n^2)) =2\exp(-\operatorname{erf}(1)))+8\exp(-4\operatorname{erf}(4)) > 1.0076 $$