Does anyone know of a closed form solution to the following integral?

Question

Does anyone know of a closed form solution to the following integral? $$ \DeclareMathOperator\erf{erf} \newcommand{d}{\;\mathrm{d}} \int^{+\infty}_{-\infty} \erf^{\;m}\!(x) \frac{\d^n \erf(x)}{\d x^n}\d x $$

Where $m$ and $n$ are integers $\geq 0$ and $\geq 1$ respectively. Integration by parts does not seem to help and I have scoured the internet to find something similar without success.

Answer 1

Partial answer:

$$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{m}{\frac {d}{dx}}{{\rm erf}\left(x\right)}{dx}= \cases{0&$m= {\it odd}$\cr 2\, \left(m+1 \right) ^{-1}&$m= even$\cr} $$

$$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{2\,n}{\frac {d^{2}}{d{x}^{2}}}{{\rm erf}\left(x\right)}{dx}=0 $$

$$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{2\,n+1}{\frac {d^{3}}{d{x}^{3}}}{{\rm erf}\left(x\right)}{dx}=0 $$

Conjecture:

$$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{n}{\frac {d^{m}}{d{x}^{m}}}{{\rm erf}\left(x\right)}{dx} =0$$ when $m$ and $n$ are odd numbers; or when $m$ and $n$ are even numbers.

Answer 2

Another partial answer:$\DeclareMathOperator\erf{erf}$

$$ \int_{-\infty}^\infty \erf(x) \dfrac{d^n}{dx^n} \erf(x)\; dx = \cases{0 & $n$ odd\cr (-1)^{n/2} 2^{(n+1)/2}\Gamma(\frac{n-1}{2})/\pi & $n$ even }$$