Why can I change the integration unit in Fourier transform>

Question

I was hoping for an explanation of the following steps in taking the Fourier transform of the Klein-Gordon equation.

\begin{equation} \phi (\boldsymbol x,t)=\int \frac{d^3p}{(2\pi)^3}e^{ip\cdot x}\phi(\boldsymbol p,t)\nonumber \end{equation} Entering this expression into the Klein-Gordon equation, \begin{equation} \int \frac{d^3p}{(2\pi)^3}e^{ip\cdot x}\bigg(\frac{\partial ^2}{\partial t^2}-\nabla ^2+m^2\bigg)\phi(\boldsymbol p,t)=\int \frac{d^3p}{(2\pi)^3}e^{ip\cdot x}\bigg(\frac{\partial ^2}{\partial t^2}-\boldsymbol p^2+m^2\bigg)\phi(\boldsymbol p,t)=0\nonumber \end{equation}

\begin{equation} \int d^3x\frac{d^3p}{(2\pi )^3}e^{i(p-p')\cdot x}\bigg(\frac{\partial ^2}{\partial t^2}+\boldsymbol p^2+m^2\bigg)\phi(\boldsymbol p,t)=0\ \ \ \ \ \ \ \ \ \ (?) \end{equation} Then $\int d^3xe^{-i(p-p')\cdot x}=\delta (p-p')$ (?), \begin{equation} \int \frac{d^3p}{(2\pi )^3}\delta (p-p')\bigg(\frac{\partial ^2}{\partial t^2}-\boldsymbol p^2+m^2\bigg)\phi(\boldsymbol p,t)=0 \end{equation} Finally by the properties of the delta function this gives,

\begin{equation} \bigg(\frac{\partial ^2}{\partial t^2}-\boldsymbol p^2+m^2\bigg)\phi(\boldsymbol p,t)=0 \end{equation} How is it possible to add in the $d^3x$ and integrate over that and why does the argument of the exponential change to $p-p'$, and finally why does $\int d^3xe^{-i(p-p')\cdot x}=\delta (p-p')$?

Answer 1

1. Multiply both sides by $\exp(-i\mathbf{p}' \cdot \mathbf{x})$ and integrate over $\mathbb{R}^3$. Swap order of integration.
2. The dirac delta result is can be found many places. See rule 303 here.

Also, you shouldn't use the same name for a function and its Fourier transform. It's confusing; use $\hat{\phi}$ or something. Also also, you mean $||\mathbf{p}||^2$ where you have $\mathbf{p}^2$. Also also also, you swapped the negative sign to a plus in your first question mark line. Also also also also, this seems like a very convoluted way of expressing how the Fourier transform transforms the Laplacian operator... this should just be a one-line statement.