This problem arose while I was teaching Statistical Quality Control at the University to Industrial Engineers, and I asked myself where do the parameters $d_{2}$ and $d_{3}$, used to compute control charts, came from? I did some research and found the article named Tables of Range and Studentized Range by H. Leon Harter https://www.jstor.org/stable/2237810?seq=1.
Page 1124 presents the following equation that was used in order to compute the moments of the range statistics.
$$ E\left(W^{k}\right) = n\left(n-1\right)\int_{-\infty}^\infty \Bigg\{ \int_{0}^\infty W^{k} \Big[ \Phi(X+W) - \Phi(X) \Big]^{n-2} \phi(X+W)dW \Bigg\}\phi(X)dx $$
With
$$ \phi(X) = (2\pi)^{-1/2}e^{-X^{2}/2} \qquad \Phi(X) = \int_{0}^X \phi(X)dX $$
The article itself describes that in order to solve the equation a Lagrangian integration formula was used for the inner integral and the trapezoidal rule for the outer internal. Also mentioned that in the case $n=2$ and $k=1$ the equation has a closed-form with result $E(W) = 2/ (\pi)^{1/2}$
When the suggested substitutions are made, $n=2$ and $k=1$ ,equation reduces to the following form:
$$ E(W) = 2\int_{-\infty}^\infty \Bigg\{ \int_{0}^\infty W (2\pi)^{-1/2}e^{\frac{-(X+W)^{2}}{2}}dW \Bigg\}(2\pi)^{-1/2}e^{-X^2/2}dX $$
I'm guessing, and it looks, that in order to solve this particular case the outer integral looks like the normal integral so, it has to be equal to $1$, if so, the inner integral must have the value of $1/(\pi)^{1/2}$ but it doesn't look like a closed-form or easy to compute.
How to obtain the closed-form for the particular case? and how can you set up the numerical methods in order to solve the equation?
I would appreciate any advice or answers
P.S. If you want to check a reference for the factor for Constructing Variables Control Charts, the book Introduction to Statistical Quality Control 6th edition by Douglas C. Montgomery on page 702 could be checked or I leave the following link http://chafra.bilkent.edu.tr/SOM/Control%20Charts%20Factors.pdf