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This may be quite simple, but I am a little rusty and have had quite a few attempts at this and would appreciate some help. The form of the integrand does not really matter, I am trying to change the measure of this double integration and also the corresponding boundaries. The integration limits are

$$\int_{0}^{t}dt_{1}\int_{0}^{t}dt_{2} $$ So essentially I have a rectangle in the $(t_{1}-t_{2})$ plane. I would like to change the integration variable to $$\int_{y_{1_{int}}}^{y_{1_{fin}}}dy_{1}\int_{y_{2_{int}}}^{y_{2_{fin}}}dy_{2} $$ where $y_{1}=\frac{t_{1}+t_{2}}{2}$ and $y_{2}=t_{2}-t_{1}$. I think I need to split the integral into two parts but I can't figure out how to do this.

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First thing first is getting our new volume element. For this we use the Jacobian, in this case that is $$\begin{align} \mathcal{J} &=\frac{\partial(y_1,y_2)}{\partial(t_1,t_2)} \\ &=\left| \begin{array}{cc} y_{1t_1} & y_{1t_2} \\ y_{2t_1} & y_{2t_2} \end{array}\right|=\left| \begin{array}{cc} 1/2 & 1/2 \\ -1 & 1 \end{array}\right| \\ &=1 \end{align}$$ So for us in this case the volume element stays the same which just makes things easier. Next is changing the bounds. Our region is the region enclosed by the lines $$\begin{array}{cc} t_1=0 & t_1=t \\ t_2=0 & t_2=t \end{array}$$ We wish to get these boundaries in terms of $y_1$ and $y_2$, so to do that we solve for $t_1$ and $t_2$ in terms of $y_1$ and $y_2$. $$\begin{align} t_1=y_1-\frac{1}{2}y_2 \\ t_2=y_1+\frac{1}{2}y_2 \end{align}$$ Thus our new boundaries are $$\begin{array}{cc} y_1-\frac{1}{2}y_2=0 & y_1-\frac{1}{2}y_2=t \\ y_1+\frac{1}{2}y_2=0 & y_1+\frac{1}{2}y_2=t \end{array}$$ Which gives us a parallelogram whose bottom vertex is the origin and whose height is $t$ and width is $2t$ (demos plot of bounds for $t=4$). So from the looks of it we need to split up the integral to get $$\begin{align} I &=\int_{-t}^0dy_2\int_{-y_2/2}^{t+y_2/2}dy_1+\int_{0}^tdy_2\int_{y_2/2}^{t-y_2/2}dy_1 \\ &=\int_{-t}^tdy_2\left(\int_{-y_2/2}^{t+y_2/2}dy_1+\int_{y_2/2}^{t-y_2/2}dy_1\right) \end{align}$$ Which I believe is the simplest way to put it.

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  • $\begingroup$ Thank you very much, that was very clearly explained. You really helped me out. $\endgroup$ – SAMCRO Jul 21 '16 at 20:15
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I found a slightly different way of doing things and thought I would share it with people.this shows a change of basis essentially, hopefully the drawing is clear

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