TITLE: Nested integral: volume of a right simplex
QUESTION [6 upvotes]: Consider the integral
$$I_n \equiv \int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots \int_{t_n = t_{n-1}}^T dt_n \, .$$
I suspect the result should be $I_n = T^n/n!$ but would like to prove it.
Doing the $n^\text{th}$ integral gives
\begin{align}
I_n &= \int_{t_1=0}^T dt_1 \cdots \int_{t_{n-1}=t_{n-2}}^T dt_{n-1} \, (T-t_{n-1}) \\
&= T I_{n-1} - \int_{t_1=0}^T dt_1 \dots \int_{t_{n-1}=t_{n-2}}^T dt_{n-1} \, t_{n-1} \, .
\end{align}
Is there some way to evaluate the second term or to come up with a useful recurrence relation or induction step?

REPLY [2 votes]: Here's an alternative, albeit perhaps a less inspiring approach. The way to read off that the integral $$I_n = \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-1}}^T dt_n$$ is measuring volume of the regular right $n$-simplex of height $T$ is to see that the domain of the integral is $V = \{(t_1, t_2, \cdots, t_n) \in \Bbb R^{n+1} : t_{k-1} \leq t_k \leq T; t_0 = 0 \}$. This is the convex subset of $\Bbb R^{n+1}$ defined by the set of linear inequalities $0 \leq t_1 \leq \cdots \leq t_n \leq T$; one can easily see that this is nothing other than a simplex (inductive argument: fix a $t_1 = t_{01}$ and observe that at that slice it's just the same convex subset in one dimension lower. It's clearly a simplex for $n = 1$.)
Here's a picture for $n = 2$ (stolen from Shifrin's "Multivariable Mathematics"):
where $t_1 = x$ and $t_2 = y$ are the variables in the convention of the image. Here we have changed the slices of the volume from $(t_1 = c)$'s for various $c \in [0, T]$ to $(t_2 = c)$'s for various $c \in [0, T]$. What this accomplishes is the following; we change the limits of our integral to $0 \leq t_2 \leq T$ and $0 \leq t_2 \leq t_1$, and accordingly switches the order of the integrals: 
$$I_2 = \int_0^T dt_1 \int_{t_1}^T dt_2 = \int_0^T dt_2 \int_0^{t_2} dt_1$$
We can similarly generalize. Change the limits (and order) of the integral $I_n$ from the inequalities $0 \leq t_1 \leq T, t_1 \leq t_2 \leq T, \cdots, t_{n-1} \leq t_n \leq T$ to $0 \leq t_n \leq T, 0 \leq t_{n-1} \leq t_n, \cdots, 0 \leq t_1 \leq t_2$. Hence, we obtain the equality $$I_n = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_2} dt_1$$
Notice that this is secretly nothing but Fubini; we have changed the order of integrals. Now, why is this useful? Just notice that we can start evaluating the integral from the innermost iterated integral
$$\begin{align} I_n = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_3} dt_2 \int_0^{t_2} dt_1 & = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_{k+1}} dt_k \; t_k^{k-1}/(k-1)! \\ &= \int_0^T dt_n \; t_n^{n-1}/(n-1)! \\ &= T^n/n!\end{align}$$
So your integral is indeed the volume of the regular right-angled $n$-simplex of height $T$ which is indeed equal to $T^n/n!$.