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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?

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2 Answers 2

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I think induction is your best bet here, and it doesn't even really require any $I_n$ notation. We can simply evaluate the integral on it's own. Starting with your first step, which you did correctly, we begin to integrate. $$\int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots \int_{t_n = t_{n-1}}^T dt_n = \int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots\int_{t_{n-1}=t_{n-2}}^T (T-t_{n-1})dt_{n-1}$$ Integrating by power rule (and chain rule for the negative), we get $$\int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots\int_{t_{n-2}=t_{n-3}}^T -\frac{(T-t_{n-1})^{2}}{2!} \Bigg|_{t_{n-2}}^{T} dt_{n-2} \\ =\cdots\int_{t_{n-2}=t_{n-3}}^T -\frac{\left(T-T\right)^2}{2!} -\left(-\frac{\left(T-t_{n-2}\right)^2}{2!}\right) dt_{n-2} \\ = \cdots\int_{t_{n-2}=t_{n-3}}^T \frac{\left(T-t_{n-2}\right)^2}{2!}dt_{n-2}$$ And finally, we see a pattern, so let us prove the induction step $$\int_{t_{k}=t_{k-1}}^T\frac{(T-t_k)^a}{a!} dt_k = -\frac{(T-t_k)^{a+1}}{(a+1)!}\Bigg|_{t_{k-1}}^T = \frac{(T-t_{k-1})^{a+1}}{(a+1)!}$$ Now that we have that, we integrate along until we get to the final integral with $t_1$ $$\int_{t_1=0}^T \frac{(T-t_1)^{n-1}}{(n-1)!}dt_1 = -\frac{(T-t_1)^n}{n!} \Bigg|_0^T = \frac{T^n}{n!}$$ And there we have the answer: the volume of a regular right simplex!

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  • $\begingroup$ I don't understand where $-(T-T)^2/2!$ comes from. I submitted an edit to remove it, but apparently it was rejected. $\endgroup$
    – DanielSank
    Commented Dec 6, 2017 at 19:18
  • $\begingroup$ It comes from the fact that we are evaluating a definite integral. I will add a line above to show this. $\endgroup$ Commented Dec 6, 2017 at 19:52
  • $\begingroup$ Oh I see. I would have integrated the $T$ and $-t_{n-1}$ separately, but your way is better. $\endgroup$
    – DanielSank
    Commented Dec 6, 2017 at 20:02
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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"):enter image description here

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 $T^n/n!$.

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