Is the parity of error function enough to show :$\int_{-l}^{l} \exp ({\operatorname{-x^2erf(x)})dx=\int_{-l}^{l} \exp({\operatorname{x^2erf}}(x)})dx$?

Question

I have tried to show the below identity using the parity of both error function and exp function but I didn't succeed, then my question here is there any analytical way to show this identity or

Is the parity of error function enough to show :$$\int_{-l}^{l} \exp (-x^2{\operatorname{erf}(x))~dx=\int_{-l}^{l} \exp(x^2{\operatorname{erf}}(x)})~dx$$ with $l$ is a real number.

Answer 1

Since $\operatorname{erf}$ is an odd function, i.e. $$\operatorname{erf}(-x)=-\operatorname{erf}(x),$$ substituting $ x=-t$ gives \begin{align*} \int_{-l}^l \exp(x^2\operatorname{erf}(x))~\mathrm dx&=\int_{l}^{-l}\exp\big((-t)^2\operatorname{erf}(-t)\big)~\mathrm d(-t)\\ &=\int_{-l}^l \exp(-t^2\operatorname{erf}(t))~\mathrm dt. \end{align*}