Change of variable formula for $n$ dimensional integral

Question

I want to use the formula for a change of variables in $n$ dimensions to compute the following integral \begin{align*} \int_0^\infty\int_{x_1}^\infty\cdots\int_{x_{n-1}}^\infty\mathbb{1}_{B_1\times B_2\times...\times B_n}(x_1, x_2-x_1,...,x_n-x_{n-1})e^{-x_n}\,dx_n...dx_1, \end{align*} where $B_1,...,B_2$ are sets in the Borel $\sigma$-algebra of $\mathbb{R}^+$. The substitution that I want to make is letting $t_1=x_1$ and $t_i=x_i-x_{i-1}$ for $2\le i\le n$. I know the answer will be \begin{align*} \int_0^\infty\int_0^\infty\cdots\int_0^\infty\mathbb{1}_{B_1\times B_2\times....\times B_n}(t_1, t_2,...,t_n)e^{-\sum_{i=1}^nt_i}\,dt_n...dt_1 \end{align*} but I want to show this rigorously, in particular, as mentioned, using the formula for a change of variables in $n$ dimensions. It's just been years since I've seen Calc III and I don't remember the formula, the only explicit formulas I can easily find are for 2 and 3 dimensions. I was wondering if someone wouldn't mind jogging my memory and working out all the details for this substitution, thanks!

Answer 1

You should be able to work this out yourself. It does not even require the multi-dimensional change of variable formula, only the original "$u$-substitution" change of variables.

For example, the inner integral in

$$\int_0^{\infty} \int_{x_1}^{\infty} f(x_1, x_2-x_1)e^{-x_2}dx_2dx_1$$

is

$$\int_{x_1}^{\infty} f(x_1, x_2 - x_1)e^{-x_2} dx_2$$

where $x_1$ is a fixed positive real number (constant). Making the change of variables $x_2 \mapsto (x_2 + x_1)$, we see that this integral is equal to

$$\int_0^{\infty} f(x_1, x_2) e^{-x_1 - x_2}dx_2$$

and therefore

$$\int_0^{\infty} \int_{x_1}^{\infty} f(x_1, x_2-x_1)e^{-x_2}dx_2dx_1 = \int_0^{\infty}\int_0^{\infty} f(x_1, x_2) e^{-x_1 - x_2}dx_2 dx_1.$$