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I am trying to show the following:

$$ \int_{B(\mathbf{\xi},\rho)} f(\mathbf{x})d\mathbf{x}=\int_{B(\mathbf{0},\rho)} f(\mathbf{x}+\mathbf{\xi})d\mathbf{x}, $$

where $B(\mathbf{\xi},\rho)$ and $B(\mathbf{0},\rho)$ are open balls of radius $\rho$ centred around $\mathbf{\xi}$ and $\mathbf{0}$, respectively. To do so, I am using the following formula:

$$ \int_R f(\mathbf{x})d\mathbf{x}=\int_T f(\Phi(\mathbf{u}))\left| \det J_\Phi(\mathbf{u}) \right|d\mathbf{u}, $$

where I have defined $T=B(\mathbf{0},\rho)$, $R=B(\mathbf{\xi},\rho)$, and $\Phi(\mathbf{u})=\mathbf{u}-\xi$. The Jacobian matrix generated by this choice of transformation is an identity matrix, and so its determinant is $1$. Thus we have:

$$ \int_{B(\mathbf{\xi},\rho)} f(\mathbf{x})d\mathbf{x}=\int_{B(\mathbf{0},\rho)} f(\mathbf{u}-\mathbf{\xi})d\mathbf{u}. $$ This is where I am stuck, I am not sure what the final step is and I feel like I am missing something obvious. Any help would be appreciated.

Cheers!

Aig
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Thomas Cairns
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    You have the transformation backwards. With $\Phi(u) = u - \xi$, the region of integration shifts to $B(2\xi,\rho)$, not $B(0,\rho)$. – Ninad Munshi Mar 07 '24 at 07:04
  • @NinadMunshi Ah, that makes much more sense. So then we have $\int_{B(\mathbf{\xi},\rho)} f(\mathbf{x})d\mathbf{x}=\int_{B(\mathbf{0},\rho)}f(\mathbf{u}+\mathbf{\xi})d\mathbf{u}$, finishing the proof? – Thomas Cairns Mar 07 '24 at 08:11
  • Yep that is pretty much it. – Ninad Munshi Mar 07 '24 at 08:15

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