Let $f$ be an integrable function from $\mathbb{R}\to\mathbb{R}.$
I have the following integral $$I=\int_{-a}^a\int_{-a}^a\int_{\mathbb{R}}f(x-y-z)dxdydz.$$
Is is possible to make a change of variable such that the integral will only depends on $(y,z)\;?$
I tried $(u,v,w)=(x,x-y,z)$ if I am not mistaken I have $$I=2a\int_{0}^{2a}\int_{\mathbb{R}}f(y-z)dydz.$$ Is it correct ?
Try $(u,v,w)=(x,y,x-y-z)$. For $(x,y,z)\in[-a,a]\times[-a,a]\times\mathbb R$ you have $(u,v,w)\in[-a,a]\times[-a,a]\times\mathbb R$ and $|J|=|\frac{\partial(x,y,z)}{\partial(u,v,w)}|=1$, so $$\int_{-a}^a dx \int_{-a}^a dy \int_{-\infty}^\infty dz f(x-y-z) = \int_{-a}^a du \int_{-a}^a dv \int_{-\infty}^\infty dw f(w) = 4a^2 \int_{-\infty}^\infty f(w)dw $$