Issue with integrating double integrals 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?
 A: You did not intergate w.r.t. $t_1$.
$\int_0^{t_2}dt_1=t_2$ and $\int_0^{t} t_2dt_2=\frac {t^{2}}2$.
A: 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 $$\int_{0<t_{1}<t_{2}<...<t_{n}<t}dt_{1}dt_{2}...dt_{n}=\frac{t^{n}}{n!}$$
