Issue with integrating double integrals

Question

I am trying to show that $\int_{0<t_1<t_2<t} dt_1dt_2$ = $\frac{t^2}{2}$?

This is my working, which I know is wrong but i am not sure why: $$\int_{0<t_1<t_2<t}dt_1dt_2=\int(\int \mathbb{1}_{0<t_1<t_2<t} \;dt_1)dt_2=\int t_2\mathbb{1}_{0<t_2<t}\;dt_2=\frac{t}{2}$$

Can someone explain to me what is wrong with my working?

The middle step should give int t_2 dt_2... – Benjamin_Gal Dec 17 '21 at 11:42 — Benjamin_Gal, Dec 17 '21 at 11:42

score 2 · Answer 1 · answered Dec 17 '21 at 11:41

2

You did not intergate w.r.t. $t_1$.

$\int_0^{t_2}dt_1=t_2$ and $\int_0^{t} t_2dt_2=\frac {t^{2}}2$.

answered Dec 17 '21 at 11:41

Kavi Rama Murthy

311,013

Ahh silly me. Thanks for pointing that out – dante Dec 17 '21 at 11:44

score 2 · Answer 2 · answered Dec 17 '21 at 11:53

A more intuitive and general way to tackle this:-

Let $$0<t_{1}<t_{2}<...<t_{n}<t$$.

Then $$\int_{0<t_{1}<t_{2}<...<t_{n}<t}dt_{1}dt_{2}...dt_{n}=\int_{[0,t]^{n}}\mathbf{1}_{\{0<t_{1}<t_{2}<...<t_{n}<t\}}dt_{1}dt_{2}...dt_{n}$$.

Now forget for a moment that $0<t_{1}<t_{2}<...<t_{n}<t$. If this condition was not there you are simply looking at the volume of an $n$-dimensional cube of side $t$.

Now $0<t_{1}<t_{2}<...<t_{n}<t$ means you are only restricting yourself to the region where this holds. But if these were to vary unconditonally you get the whole region.

Now there are precisely $n!$ ways of arranging $t_{1},t_{2},...,t_{n}$ in ascending order. And the sum of all those $n!$ volumes equal to the volume of the entire $n-$cube. By symmetry each will have the same volume say $V$.

Then $$n!V=t^{n}$$. So $$V=\frac{t^{n}}{n!}$$.

Hence $$\int_{0<t_{1}<t_{2}<...<t_{n}<t}dt_{1}dt_{2}...dt_{n}=\frac{t^{n}}{n!}$$

Issue with integrating double integrals

2 Answers2