My question is about this integral: $\iiint {e^{i\overrightarrow{k}\overrightarrow{x}} }\frac{1} {k^2} dk^3$
I'm not interested to the solution but how it's solved in spherical coordinates. In fact, it's usually written in spherical coordinates as follows:
$$ \iiint e^ {ikx \cos\theta}sin\theta dk d\phi d\theta$$
My question is about the exponent which I cannot understand: The $cos\theta$ comes from the dot product of $ \overrightarrow{x}$ and $\overrightarrow{k}$ . How this angle, which is the angle between $ \overrightarrow{x}$ and $\overrightarrow{k}$, should be the same of the polar angle that is the variable of integration? Its definition is different too; in fact it's the angle between the z-axis and the vector.
For any fixed $\mathbf{x}$ with $x=|\mathbf{x}|$ there is always a change of variable $T$, orthogonal rotation, so that $x\mathbf{e}_3=T\mathbf{x}$ or $T^t(x\mathbf{e}_3) = \mathbf{x}$. Hence it follows \begin{align} \iiint \frac{e^{i\mathbf{x}\cdot\mathbf{k}}}{k^2}\ d^3\mathbf{k} =&\ \iiint \frac{e^{iT^t(x\mathbf{e}_3)\cdot\mathbf{k}}}{k^2}\ d^3\mathbf{k}\\ =&\ \iiint \frac{e^{ix\mathbf{e}_3\cdot T(\mathbf{k})}}{k^2}\ d^3\mathbf{k}\\ \text{change of coordinates } =&\ \iiint \frac{e^{ix\mathbf{e}_3\cdot \mathbf{k}'}}{k^2}\ d^3\mathbf{k'} = \iiint \frac{e^{ixk'_3}}{k^2}\ d^3\mathbf{k'} \end{align} where $\mathbf{k}' = T\mathbf{k}$. Note that $\mathbf{k}'\cdot\mathbf{k}' = \mathbf{k}\cdot \mathbf{k} = k^2$. Now use spherical coordinates.