Integration of $F(\sum_k x_k)$ over positive orthant 
Problem
Suppose we integrate some function $F\left(\sum\limits_{k=1}^n x_k\right)$ over the positive orthant $[0,\infty)^n$. Show that this this is proportional to the integral $\int\limits_0^\infty s^{n-1}F(s)\,ds$. What is the constant of proportionality? If this is a well-known result, a reference would be appreciated.

Motivation
Suppose I want to solve the inhomogeneous 1st-order ODE $(1-D)f(x)=g(x)$. Then
$$D(e^{-x} f)=e^{-x}(f(x)-f'(x))=e^{-x}(1-D)f(x)=e^{-x}g(x),$$ so if I integrate both sides over $[x,\infty)$ (and assume that $e^{-x}f(x)$ vanishes at infinity) I can obtain the integral representation
$$f(x)=(1-D)^{-1}g(x)=\int_{x}^\infty dx'\,e^{-(x'-x)}g(x')=\int_{0}^\infty ds\,e^{-s}g(s+x) \quad\quad(x'=s+x).$$
If I instead have the 2nd-order ODE $(1-D)^2 f(x)=g(x)$, then I will instead obtain the double-integral representation
\begin{align}
f(x)&=(1-D)^{-2}g(x)\\&=(1-D)^{-1}\int_{0}^\infty ds\,e^{-s}g(s+x)\\&=\int_{0}^\infty \int_{0}^\infty ds' ds\,e^{-(s+s')}g(s+s'+x)
\end{align}
To make this more convenient, I can map $(s,s')\mapsto(\frac{s+s'}{2},\frac{s-s'}{2})$ and modify the bounds of integration accordingly to obtain
$$f(x)=(1-D)^{-2}g(x)=\frac{1}{2}\int_0^\infty ds\,e^{-s}g(s+x)\int_{-s}^{s} ds'=\int_0^\infty ds\,se^{-s}g(s+x).$$
So acting twice with $(1-D)^{-1}$ merely introduces a linear factor into the integration over $s$. This surely generalizes to $n$ applications of $(1-D)^{-1}$ (presumably by introducing a factor of $s^{n-1}$ instead) but I'm unfamiliar with such an identity. Can anyone supply a proof/reference?
 A: This is a special case of Type I Dirichlet integrals:
$$\int_{[0,\infty)^n} f\left(\sum_{k=1}^n t_k\right) \prod_{k=1}^n t_k^{\alpha_k-1} \prod_{k=1}^n dt_k 
= \frac{\prod_{k=1}^n\Gamma(\alpha_k)}{\Gamma\left(\sum_{k=1}^n \alpha_k\right)}
\int_0^\infty f(\tau) \tau^{(\sum_{k=1}^n \alpha_k) - 1} d\tau
\tag{*1}$$
When all $a_k = 1$, this reduces to
$$
\int_{[0,\infty)^n} f\left(\sum_{k=1}^n t_k\right) \prod_{k=1}^n dt_k 
= \frac{1}{(n-1)!} \int_0^\infty f(\tau) \tau^{n - 1} d\tau
$$
and the proportional factor you seek is $\frac{1}{(n-1)!}$.
Update
To prove the formula, let us introduce following change of coordinates
$$[0,\infty)^n \ni ( t_1, t_2, \ldots, t_n ) \quad\leftrightarrow\quad (\tau, \lambda_1, \lambda_2, \ldots, \lambda_{n-1}) \in [0,\infty) \times [0,1]^{n-1}$$
defined by the relations
$$\begin{array}{rcl}
t_1 + t_2 + t_3 + \cdots + t_n &=& \tau\\
t_2 + t_3 + \cdots + t_n &=& \tau \lambda_1\\
t_3 + \cdots + t_n &=& \tau \lambda_1\lambda_2\\
&\vdots&\\
t_n &=& \tau\lambda_1\lambda_2\cdots\lambda_{n-1}
\end{array}$$
In the new coordinates, the hypervolume element becomes
$$\begin{align}
\bigwedge_{k=1}^n d t_k
&= d(t_1 + t_2 + \cdots + t_n) \wedge d(t_2+\cdots+t_n) \wedge \cdots \wedge d t_n\\
&= d\tau \wedge d(\tau\lambda_1) \wedge \cdots \wedge d(\tau\lambda_1\cdots\lambda_{n-1})\\
&= \tau^{n-1} \prod_{k=1}^{n-1} \lambda_k^{n-1-k} d\tau \wedge \bigwedge_{k=1}^{n-1} d\lambda_k
\end{align}
$$
Notice
$$\begin{align}
  \prod_{k=1}^n t_k^{\alpha_k-1}
= & \prod_{k=1}^{n-1} \left( \tau(1 - \lambda_k)\prod_{j=1}^{k-1}\lambda_j\right)^{\alpha_k-1}
\times \left(\tau\prod_{j=1}^{n-1} \lambda_j \right)^{\alpha_n-1}\\
= & \tau^{(\sum_{j=1}^n \alpha_j) - n}
    \prod_{k=1}^{n-1} (1-\lambda_k)^{\alpha_k-1} \lambda_k^{(\sum_{j=k+1}^n\alpha_j) - (n-k)}
\end{align}
$$
The integral in LHS$(*1)$ can be rewritten as
$$C \int_0^\infty f(\tau)\tau^{(\sum_{k=1}^n \alpha_k) - 1} d\tau\tag{*2}$$
and the proportional constant $C$ is given by
$$\begin{align}
C 
&=\prod_{k=1}^{n-1}\int_0^1 (1-\lambda_k)^{\alpha_k-1}\lambda_k^{(\sum_{j=k+1}^n\alpha_j)-1} d\lambda_k\\
&= \prod_{k=1}^{n-1}\frac{\Gamma(\alpha_k)\Gamma(\sum_{j=k+1}^n\alpha_j)}{\Gamma(\sum_{j=k}^n \alpha_j)}\\
&= \prod_{k=1}^{n-1} \Gamma(\alpha_k) \times 
\prod_{k=1}^{n-1}\frac{\Gamma(\sum_{j=k+1}^n\alpha_j)}{\Gamma(\sum_{j=k}^n\alpha_j)}
\\
&= \prod_{k=1}^{n-1} \Gamma(\alpha_k) \times 
\frac{\Gamma(\alpha_n)}{\Gamma(\sum_{j=1}^n\alpha_j)}\\
&= \frac{\prod_{k=1}^n \Gamma(\alpha_k)}{\Gamma(\sum_{k=1}^n \alpha_k)}
\end{align}
$$
Substitute this expression of $C$ into $(*2)$ reproduces RHS$(*1)$.
A: Derivation #1
Let us introduce the integral operator $\mathcal{L}[\alpha]$ for real $\alpha\geq 1$, acting as 
$$\mathcal{L}[\alpha]=\frac{1}{\Gamma(\alpha)}\int_0^\infty ds\,s^{\alpha-1}e^{-(k-D_z)s}.$$
We can prove that these operators possess the semigroup property $\mathcal{L}[\alpha]\mathcal{L}[\beta]=\mathcal{L}[\alpha+\beta]$. Observe that 
\begin{align}
\mathcal{L}[\alpha]\mathcal{L}[\beta]
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(\beta)}\int_0^\infty \int_0^\infty ds_1 ds_2\,s_1^{\beta-1}s_2^{\alpha-1}e^{-(s_1+s_2)(k-D_z)}\\
\end{align}
We simplify this integration  by the substitutions $s:=s_1+s_2$, $t:=s_1/s$. Accounting for the change of measure and the bounds of integration, we then have the factorization
$$
\mathcal{L}[\alpha]\mathcal{L}[\beta]F(z)=\int_0^\infty dt\,\frac{t^{\beta-1}(1-t)^{\alpha-1}}{\Gamma(\alpha)\Gamma(\beta)}\times 
\int_0^\infty  ds \, s^{\alpha+\beta-1}e^{-s(k-D_z)}
$$
But we recognize the first integral as the definition of the Beta function, which has the well-known representation in terms of the Gamma function as $\text{B}(\alpha,\beta)=\dfrac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$. Consequently as claimed we have
$$
\mathcal{L}[\alpha]\mathcal{L}[\beta]=\dfrac{1}{\Gamma(\alpha+\beta)}
\int_0^\infty  ds \, s^{\alpha+\beta-1}e^{-s(k-D_z)}=\mathcal{L}[\alpha+\beta]
.$$
This semigroup property immediately generalizes to any such partition: If $\alpha=\sum_{j=1}^n \alpha_j$ with $\alpha_j\geq 1$, then $\mathcal{L}\bigl[\alpha]=\prod_{j=1}^n\mathcal{L}[\alpha_j]$.
Applying this to a generic test function $F(z)$
yields \begin{align}
\left(\prod_{j=1}^n\mathcal{L}[\alpha_j]\right)F(z)&=\int_{\mathbb{R}_+^n} dV_s \left(\prod_j \frac{s_j^{\alpha_j-1}e^{-k s_j}}{\Gamma(\alpha_j)}\right)g\left(z+\sum_j s_j\right)\\
=\mathcal{L}[\alpha]F(z)&=\frac{1}{\Gamma(\alpha)}\int_0^\infty ds\,s^{\alpha-1}e^{-k s}F(z+s)\end{align}
Finally, the limiting case $(k,\alpha_j,z)\to(0,1,0)$
establishes the identity requested in the question, and gives the constant of proportionality as $\Gamma(n)^{-1}$.
Derivation #2
Proceeding along the lines set out in the motivation given in my question, here is a purely formal approach (which is to say, I won't worry about rigor). If we modify the initial 1st-order ODE slightly to $(1-\tau-D)f=g$, then we may proceed as above to find
$$f(x)=(1-\tau-D)^{-1}g(x)=\int_0^\infty ds\,e^{-(1-\tau)s}g(s+x),$$ and rearranging this a bit gives
$$(1-\tau-D)^{-1}g(x)=\frac{(1-D)^{-1}}{1-\tau (1-D)^{-1}}g(x)=\int_0^\infty ds\,e^{\tau s}e^{-s}g(s+x).$$
Identifying coefficients of $\tau^{n-1}$ on both sides then yields
$$(1-D)^{-n}g(x)=\dfrac{1}{(n-1)!}\int_0^\infty ds\,e^{-s}s^{n-1}g(s+x).$$
But we can also directly use $(1-D)^{-1} g(x)=\int_0^\infty ds\, e^{-s}g(s+x)$ a total of $n$ times, which results in an integral over the positive orthant $R_+^n$:
$$(1-D)^{-n} g(x)=\int_{R_+^n} dV\, \exp\left(-\sum_k s_k\right)g\left(\sum_k s_k+x\right).$$ Writing $F(s)=e^{-s} g(s+x)$ and comparing the two results, we conclude 
$$\boxed{\displaystyle \int_{R_+^n}dV\,F\left(\sum_k s_k\right)=\frac{1}{(n-1)!}\int_0^\infty ds\, s^{n-1} F(s)}.$$
