Transforming a multiple intergral

Question

The following equality seems to hold: \begin{equation} \int_0^\infty\cdots\int_0^\infty F(\sum_{i=1}^kx_i)\text{ d}x_k\cdots\text{ d}x_1=\int_0^\infty \frac{y^{k-1}}{(k-1)!}F(y)\text{ d}y. \end{equation} Is there an easy way to see this?

Answer 1

As Daniel Fischer suggested, let's do the change of variables: $$ x_k = y \cdot u_k $$ where $0 < u_k<1$ such that $\sum_{k=1}^n u_k = 1$ and $y > 0$. Then Jacobian of this change of variables is $$ J = y^{n-1} $$ Hence $$ \int_0^\infty \cdots \int_0^\infty F\left(\sum_{k=1}^n x_k\right) \mathrm{d}x_1 \cdots \mathrm{d}x_n = \int_0^\infty y^{n-1} F\left(y \right) \mathrm{d}y \cdot \underbrace{\int_{u_k > 0, u_1+\cdots+u_{n-1} \leqslant 1} \mathrm{d}u_1 \cdots \mathrm{d}u_{n-1}}_{\Omega_{n-1}} $$ In order to evaluate $\Omega_{n-1}$ it is best to proceed by recursion. By making another change of variables: $$ u_{n-1} = 1 - v_{n-1} \quad u_{k} = v_{n-1} v_{k} \quad \text{for } 1 \leqslant k < n-1 $$ with Jacobian $J = \left(v_{n-1}\right)^{n-2}$ we have $$ \Omega_{n-1} = \underbrace{\int_0^1 v_{n-1}^{n-2} \mathrm{d} v_{n-1}}_{\frac{1}{n-1}} \cdot \underbrace{\int_{v_k >0, v_1+\cdots+v_{n-2} < 1} \mathrm{d}v_1 \cdots \mathrm{d}v_{n-2}}_{\Omega_{n-2}} $$ Hence $$ \Omega_{n-1} = \frac{1}{n-1} \Omega_{n-2} $$ which is easy to solve: $$ \Omega_{n-1} = \frac{1}{n-1}\cdot \frac{1}{n-2} \cdots \frac{1}{2} \Omega_1 = \frac{1}{(n-1)!} $$

Answer 2

$\displaystyle{\large% \int_{0}^{\infty}\cdots\int_{0}^{\infty} {\rm F}\left(\sum_{i = 1}^{k}x_{i}\right){\rm d}x_{k}\ldots {\rm d}x_{1} = \int_{0}^{\infty}{y^{k - 1} \over \left(k - 1\right)!}{\rm F}\left(y\right)\, {\rm d}y }$

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\begin{align} & \int_{0}^{\infty}\cdots\int_{0}^{\infty} {\rm F}\,\left(\sum_{i = 1}^{k}x_{i}\right){\rm d}x_{k}\ldots {\rm d}x_{1} = \int_{0}^{\infty}\cdots\int_{0}^{\infty} \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\, \tilde{\rm F}\left(s\right) \exp\left(s\sum_{i = 1}^{k}x_{i}\right)\,{\rm d}x_{k}\ldots{\rm d}x_{1} \\[3mm]&= \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\, \tilde{\rm F}\left(s\right) \left[% \int_{0}^{\infty}\exp\left(sx\right)\,{\rm d}x \right]^{k} = \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\,\tilde{\rm F}\left(s\right) \,{\left(-1\right)^{k + 1} \over s^{k}} \\[3mm]&= \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\,\int_{0}^{\infty}{\rm d}y\,{\rm e}^{-sy}\, {\rm F}\left(y\right)\,{\left(-1\right)^{k + 1} \over s^{k}} = \int_{0}^{\infty}{\rm d}y\,{\rm F}\left(y\right)\left(-1\right)^{k + 1} \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\, {{\rm e}^{-sy} \over s^{k}} \\[3mm]&= \int_{0}^{\infty}{\rm d}y\,{\rm F}\left(y\right)\left(-1\right)^{k + 1} \int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\, {1 \over s^{k}} \sum_{n = 0}^{\infty}{\left(-y\right)^{n} \over n!}\,s^{n} \\[3mm]&= \int_{0}^{\infty}{\rm d}y\,{\rm F}\left(y\right)\sum_{n = 0}^{\infty} \left(-1\right)^{k + 1}{\left(-y\right)^{n} \over n!}\quad \overbrace{\int_{\gamma - {\rm i}\infty}^{\gamma + {\rm i}\infty} {{\rm d}s \over 2\pi{\rm i}}\,{1 \over s^{k - n}}}^{{\LARGE\delta}_{k - n,1}} = \color{#ff0000}{\large% \int_{0}^{\infty}{y^{k - 1} \over \left(k - 1\right)!}\,{\rm F}\left(y\right)} \end{align}