Multiple integral involving exponential function Working on asymptotic expansions I've found this multiple integral
$$\Phi_n(\alpha,m)=\int_{\alpha}^{\infty}e^{-x_n^2}\int_{x_{n}}^{\infty}e^{-x_{n-1}^2}\int_{x_{n-1}}^{\infty}e^{-x_{n-2}^2}\ldots\int_{x_{3}}^{\infty}e^{-x_2^2}\int_{x_{2}}^{\infty}e^{-x_{1}^2}x_1^{2m+1}\,dx_1\,dx_2\dots dx_n$$
I've been able to solve the few first cases

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*$n=1\hspace{1cm}\displaystyle\Phi_1(\alpha,m)=\int_{\alpha}^{\infty}e^{-x_1^2}x_1^{2m+1}\,dx_1=\frac{\Gamma(m+1,\alpha^2)}{2}$

*$n=2\hspace{1cm}\displaystyle\Phi_2(\alpha,m)=\int_{\alpha}^{\infty}\hspace{-0.4cm}e^{-x_2^2}\int_{x_{2}}^{\infty}\hspace{-0.4cm}e^{-x_{1}^2}x_1^{2m+1}\,dx_1\,dx_2=\int_{\alpha}^{\infty}\hspace{-0.4cm}e^{-x_2^2}\frac{\Gamma(m+1,x_2^2)}{2}\,dx_2\\
\hspace{9.77cm}=\displaystyle\sum_{k=0}^m\frac{m!\,2^{-k-1/2}}{k!}\Gamma\left(\frac{1}{2}+k,2\alpha^2\right)$
I wonder if it's possible find a closed form, perhaps in terms of any special function, of this multiple integral.
 A: To motivate ourselves, consider the following problem:
select $k$ Gaussian random variables $X_{1},\cdots,X_{k}$, such that $P(X)=\frac{1}{\sqrt{\pi}}e^{-X^{2}}$, what is the probability that they are all between $\alpha$ and $\beta$?
clearly, this object $A(\alpha,\beta,k)=A(\alpha,\beta,1)^{k}$, since the events are independent. Note that $A(\alpha,\beta,1)=\int_{\alpha}^{\beta}P(x)dx$.
If we ask what the probability that $\alpha<X_{1}<\cdots<X_{k}<\beta$, this is just a random selection as before, but it happened to be sorted, hence its probability is reduced by a factor of $\frac{1}{k!}$ due to combinatorics.
The analytic expression for this is
$$\int_{\alpha}^{\beta}P(x_{1})dx_{1}\int_{x_{1}}^{\beta}P(x_{2})dx_{2}\cdots\int_{k-1}^{\beta}P(x_{k})dx_{k}=\frac{1}{k!}A(\alpha,\beta,k)=\frac{1}{k!}A^{k}(\alpha,\beta,1)$$
To address the $x^{2m+1}$ issue, we can integrate over $\beta$ , so after returning the $\pi's$:
$$\Phi_{n}(\alpha,m)=\frac{\pi^{\frac{n}{2}}}{(n-1)!}\int_{\alpha}^{\infty}d\beta P(\beta)\beta^{2m+1}A^{n-1}(\alpha,\beta,1)$$
While this might be integrated directly, it seems that the large $n$ limit is amenable to asymptotic analysis. I have not tested the $n=1,2$ case, so I may have miscalculated, but I hope the analysis is solid.
Replacing $\beta^{2m+1}$ by its expression in terms of $A$, that is $\beta(A) = \mathrm{erf^{-1}(}A+\mathrm{\mathrm{erf}(\alpha))}$, and noting that $P(\beta)d\beta = dA$ will yield an series expansion solution, once one expands $\beta$ in terms of $A$.
