A simple calculation about distribution in the plane $\mathbb{R}^2$. Let $\Omega = \lbrace (t,x)\in\mathbb{R}^2\mid t>|x| \rbrace$ and $T\in\mathcal{D}'(\mathbb{R}^2)$ defined by $T=\partial_t\mathbb{1}_\Omega - \partial_x\mathbb{1}_\Omega$

Prove that 
  $$<T,\varphi > = 2\int_0^{+\infty}\varphi (y,y) \mathrm{d}y\quad\quad \forall \varphi\in\mathcal{S}(\mathbb{R}^2).$$

Let $\varphi\in\mathcal{S}(\mathbb{R}^2)$,
\begin{align*} 
<T,\varphi > &=  <(\partial_t -\partial_x) \mathbb{1}_\Omega,\varphi > \\ 
&=  <\mathbb{1}_\Omega,(\partial_x -\partial_t)\varphi > \\
&= \int_\Omega (\partial_x -\partial_t)\varphi (t,x) \mathrm{d}t\mathrm{d}x \\
&= \int_{-\infty}^0 \left( \int _{-x}^{+\infty}(\partial_x -\partial_t)\varphi (t,x) \mathrm{d}t\right) \mathrm{d}x + \int^{+\infty}_0 \left( \int _{x}^{+\infty}(\partial_x -\partial_t)\varphi (t,x) \mathrm{d}t\right) \mathrm{d}x
\end{align*}
But here I don't see what can I do more.
Any help would be greatly appreciated.
 A: Your life may be easier if you do not split the integral: let's write it as $\int_0^{\infty} \int_{-t}^t \, dx \, dt$ instead.
$$ \langle T , \varphi \rangle = \dots = \int_0^{\infty} \int_{-t}^t (\partial_x-\partial_t) \varphi(t,x) \, dx \, dt. $$
We have two terms here. In the first one, we can use the fundamental theorem of calculus to write down the contribution:
$$ \int_0^{\infty} \int_{-t}^t \partial_x \varphi(t,x) \, dx \, dt = \int_0^{\infty} \left[ \varphi(t,x) \right]_{-t}^t \, dt = \int_0^{\infty} \left( \varphi(t,t)-\varphi(t,-t) \right) \, dt. $$
For the other term, we have Leibniz's Rule:
$$ \partial_t \int_{a(t)}^{b(t)} f(t,x) \, dx = f(t,b(t))b'(t)-f(t,a(t))a'(t) + \int_{a(t)}^{b(t)} \partial_t f(t,x) \, dx, $$
which in our case reduces to
$$ -\partial_t \int_{-t}^{t} \varphi(t,x) \, dx = -\varphi(t,t)-\varphi(t,-t) - \int_{-t}^{t} \partial_t \varphi(t,x) \, dx, $$
so
$$ -\int_0^{\infty}\int_{-t}^{t} \partial_t \varphi(t,x) \, dx \, dt = \int_0^{\infty} (\varphi(t,t)+\varphi(t,-t)) \, dt - \int_0^{\infty} \partial_t \int_{-t}^{t} \varphi(t,x) \, dx \, dt. $$
Adding on the first term, we are down to
$$ \langle T,\varphi \rangle = 2\int_0^{\infty} \varphi(t,t) \, dt -\left[ \int_{-t}^t \varphi(t,x) \, dx \right]_{t=0}^{\infty}. $$
Now, the last term vanishes: the bottom limit is zero because the integral vanishes, and the top limit is zero because $\varphi$ is a Schwarz function, so it decays faster than any polynomial at infinity in both $t$ and $x$, and so the integral tends to zero.
