Solving an integral How to obtain an explicite formula for the following integral?
$$f_m(x_m)= \int_0^\infty\cdots \int_0^\infty x_1(x_1+x_2) \cdots (x_1+x_2+\cdots+x_m) e^{-\frac{(x_1+x_2+\cdots+x_m)^2}{2}} dx_1 dx_2 \cdots d x_{m-1},$$
where $m\geq 2$. We know that $f_2(x_2)= \int_{x_2}^\infty e^{-\frac{y^2}{2}} d y$, $f_3(x_3)= \frac{3}{2}\left(-x_3 e^{-\frac{x_3^2}{2}}+(1+x_3^2) f_2(x_3)\right)$, $f_4(x_4)= \frac{5}{16}\left(-2x_4(x_4^2+5) e^{-\frac{x_4^2}{2}}+ 2(x_4^4+6x_4^2+3) f_2(x_4)\right)$ and etc.  
 A: It is convenient to consider the function $g_n(x) = f_{n+2}(x)$ with $n \geq 0$. Before the computation, we introduce some notations:


*

*$\Delta^n(a) = \{ (x_1, \cdots, x_n) \in \Bbb{R}^n : 0 < x_1 < \cdots < x_n < a \}$.

*$I_n(x)$ for $n \geq 0$ is defined recursively by $I_0 (x) = 1$ and $I_{n+1}(x) = \int_{0}^{x} t I_n(t) \, dt$. It is easy to check that for $x > 0$,
$$ I_n(x) = \frac{x^{2n}}{2^n \cdot n!} = \int_{\Delta^n(x)} t_1\cdots t_n \, dt_1 \cdots dt_n. $$


Now let us return to the original problem and compute $g_n(x)$. With the substitution
\begin{align*}
y_1 &= x_1 \\
y_2 &= x_1 + x_2 = y_1 + x_2 \\
\vdots & \qquad \qquad \vdots \\
y_{n+1} &= x_1 + \cdots + x_{n+1} = y_n + x_{n+1}
\end{align*}
and $y = y_{n+1}$, we have
\begin{align*}
g_n(x)
&= \int_{0}^{\infty} \int_{\Delta^n (y_{n+1})} y_1 \cdots y_n y_{n+1}(y_{n+1} + x) e^{-\frac{(y_{n+1}+x)^2}{2}} \, dy_1\cdots dy_{n+1} \\
&= \int_{0}^{\infty} I_n(y) y(y + x) e^{-\frac{(y+x)^2}{2}} \, dy \\
&= \frac{1}{2^n \cdot n!} \int_{0}^{\infty} y^{2n+1}(y + x) e^{-\frac{(y+x)^2}{2}} \, dy \\
&= \frac{2n+1}{2^n \cdot n!} \int_{0}^{\infty} y^{2n} e^{-\frac{(y+x)^2}{2}} \, dy.
\end{align*}
So in principle we can compute every $g_n(x)$ by applying integration by parts several times.
