double integration change of variable

Question

I have $\int_{-T}^{T}\int_{-T}^{T} dt_1 dt_2 R(t_1-t_2)$, the book I read says it can transform into $\int_{-2T}^{2T} d \tau R(\tau) (2T - |\tau|)$ by changing variables to $\tau = t_1 - t_2$, $t= t_1$. But I cannot see it. Can you tell me why? Thanks.

Answer 1

The integral you are doing is a 2-D integral that in the variables $t_1, t_2$ is computed inside the square $\Omega$ with vertices at $(-T,-T)$, $(-T,T)$, $(T,-T)$, $(T,T)$:

$$I=\iint_\Omega dt_1\,dt_2\,R(t_1-t_2)\,.$$

When you do the change of variables to the new variables $t=t_1$ and $\tau=t_1-t_2$, the Jacobian of the transformation is $1$, and the new region of integration $\Sigma$ is the inside of the parallelogram with vertices at $(-T,0)$, $(-T,-2T)$, $(T,0)$ and $(T,2T)$.

$$I=\iint_\Sigma dt\,d\tau\,R(\tau)\,.$$

To do this integral do first the integration on the $t$ variable by splitting $\Sigma$ into two triangles $\Sigma=\Sigma_1\bigcup\Sigma_2$, with $\Sigma_1$ the triangle with $\tau<0$ and $\Sigma_2$ the triangle with $\tau\ge0$.

$$I=\iint_\Sigma dt\,d\tau\,R(\tau)=\iint_{\Sigma_1} d\tau\,dt\,R(\tau)+\iint_{\Sigma_2} d\tau\,dt\,R(\tau)\,.\qquad (1)$$

The integral ove $\Sigma_1$ is:

$$\iint_{\Sigma_1} d\tau\,dt\,R(\tau)=\int_{-2T}^0d\tau\,R(\tau)\int_{-T}^{\tau+T}dt=\int_{-2T}^0 d\tau\,R(\tau)(2T+\tau)\,.$$

Because in this integral $\tau<0$ we can replace $\tau$ with $-|\tau|$, so:

$$\iint_{\Sigma_1} d\tau\,dt\,R(\tau)=\int_{-2T}^0 d\tau\,R(\tau)(2T-|\tau|)\,.\qquad (2)$$

The integration over the second triangle $\Sigma_2$ is:

$$\iint_{\Sigma_2} d\tau\,dt\,R(\tau)=\int_0^{2T}d\tau\,R(\tau)\int_{\tau-T}^Tdt=\int_0^{2T} d\tau\,R(\tau)(2T-\tau)\,.$$

Because in this integral $\tau>0$ we can use $\tau=|\tau|$ and write:

$$\iint_{\Sigma_2} d\tau\,dt\,R(\tau)=\int_0^{2T} d\tau\,R(\tau)(2T-|\tau|)\,.\qquad (3)$$

Using the results of equations $(2)$ and $(3)$ in equation $(1)$ gives the desired result:

$$I=\int_{-2T}^0 d\tau\,R(\tau)(2T-|\tau|)+\int_0^{2T} d\tau\,R(\tau)(2T-|\tau|) = \int_{-2T}^{2T} d\tau\,R(\tau)(2T-|\tau|)$$