Solving hypersurface integral Can you help me to compute the following integral over a hypersurface in $\mathbb R^4$?
$$\int_{x_1^2 + x_2^2 - x_3^2 - x_4^2 = 1} ||x||^{-2s} d\mu$$
So here we have
$$||x||^2 := x_1^2 + x_2^2 + x_3^2 + x_4^2.$$
The measure $d\mu$ is the inherited Lebesgue messure from $\mathbb R^4$. So how do I proceed? Parametrize the hypersurface and then use appropriate formulas involving the first fundamental form?
I'm not only interested in the value of the integral but also for which $s \in \mathbb C$ it exists. I would guess for ${\rm Re}(s)>1$.
 A: Yes, you can follow the process you proposed: parametrize the surface, calculate the hypervolume form, then integrate.
The parametrization
I'll include some of my thought process.  Let the hypersurface be called $S$.  Intersecting $S$ with the planes $x_3=h$ and $x_4 = k$ gives a circle with equation $x_1^2 +x_2^2 = 1 + h^2 + k^2$.  Intersecting with planes $x_1 = h$ and $x_2 = k$ gives a circle with equation $x_3^2 + x_4^2 = h^2+k^2 - 1$.  And intersecting with the plane $x_2 = x_4 = 0$ gives a hyperbola $x_1^2-x_3^2 = 1$.  So I thought about parametrizing the two circles and the hyperbola separately.
I know that to parametrize a circle of radius $r$ in the $xy$-plane, $x = r\cos\theta$ and $y=r \sin\theta$ will work.  We can do this in the $x_1x_2$-plane, and again in the $x_3x_4$-plane.  The two radii
lie on the hyperbola $r_1^2-r_2^2=1$.  There are many ways to parametrize the hyperbola; I chose $\psi \mapsto (\sec\psi,\tan\psi)$ since I already had trigonometric functions of the other coordinates.  I suppose $\cosh$ and $\sinh$ would also work, and might be a bit cleaner.
Based on this, we see that the coordinate functions
\begin{align*}
    x_1 &= \sec\psi \cos\theta \\
    x_2 &= \sec\psi \sin\theta \\
    x_3 &= \tan\psi \cos\phi \\
    x_4 &= \tan\psi \sin\phi
\end{align*}
land in $S$.  We still need to know the domain on which this function is bijective.
Given $(x_1,x_2,x_3,x_4) \in S$, let $r_1 = \sqrt{x_1^2+x_2^2}$ and $r_2 = \sqrt{x_3^2 + x_4^2}$. Then both $r_1$ and $r_2$ are nonnegative, and $r_1^2-r_2^2 = 1$.  So there exists a unique $\psi$ with $0 \leq \psi < \frac{\pi}{2}$ such that $r_1 = \sec\psi$ and $r_2 = \tan\psi$.
Since $r_1^2-r_2^2 =1$, $r_1 > 0$, and so there exists a unique $\theta$ such that $-\pi < \theta \leq \pi$ and
$$
    \cos\theta = \frac{x_1}{r_1}
    \qquad
    \sin\theta = \frac{x_2}{r_1}
$$
Now $r_2 > 0$ is not guaranteed, but if it is true, there exists a unique $\phi$ such that
$$
    \cos\phi  = \frac{x_3}{r_2}
    \qquad
    \sin\phi  = \frac{x_4}{r_2}
$$
So if we restrict to the intervals $-\pi \leq \theta\leq \pi$, $-\pi\leq\phi\leq\pi$, $0 \leq \psi < \frac{\pi}{2}$, the mapping will be surjective, and only fail to be injective on the domain's boundary.  This will be good enough.
The integrand
The matrix of partial derivatives of this parametrization is:
$$
   J = \frac{\partial(x_1,x_2,x_3,x_4)}{\partial(\psi,\theta,\phi)}
     = \begin{bmatrix}
           \sec\psi\tan\psi\cos\theta
                & \sec\psi\tan\psi\sin\theta
                    & \sec^2\psi \cos\phi
                         & \sec^2\psi \sin\phi \\
           -\sec\psi\sin\theta
                & \sec\psi\cos\theta & 0 & 0 \\
           0 & 0 & -\tan\psi\sin\phi & \tan\psi\cos\phi
       \end{bmatrix}^T
$$
In these coordinates, $dS = \sqrt{\det J^TJ}\,d\psi\,d\theta\,d\phi$. Since
$$
    J^TJ = \begin{bmatrix}
    \sec^2\psi\tan^2\psi + \sec^4\psi & 0 & 0 \\
    0 & \sec^2\psi & 0 \\
    0 & 0 & \tan^2\psi \end{bmatrix}
$$
So
$$
    dS = \sec^2 \psi \tan \psi \sqrt{\sec^2\psi + \tan^2 \psi}\,d\psi\,d\theta\,d\phi
$$
($\tan\psi \geq 0$ since $0 \leq \psi < π/2$.)
In these coordinates, the function to be integrated is:
\begin{align*} 
    f(x_1,x_2,x_3,x_4) &= \left\Vert x \right\Vert^{-2s}
    \\&= \left[\left(x_1^2 + x_2^2 + x_3^2 + x_4^2\right)^{1/2}\right]^{-2s}
    \\&= \left(\sec^2\psi + \tan^2\psi\right)^{-s}
\end{align*}
The integral
Putting these together, we see that:
\begin{align*}
    I &= \iiint_S f\,dS
    \\&= \int_{-π}^π \int_{-π}^π \int_0^{π/2} \sec^2\psi \tan \psi  \left(\sec^2\psi + \tan^2\psi\right)^{1/2-s} \,d\psi\,d\theta\,d\phi
    \\&= 4π^2 \int_0^{π/2} \sec^2\psi \tan \psi  \left(\sec^2\psi + \tan^2\psi\right)^{1/2-s} \,d\psi
\end{align*}
Let $v = \sec^2\psi + \tan^2 \psi$, so that $dv = 4\tan \psi \sec^2\psi\,d\psi$.  Then
$$
    I = π^2\int_1^\infty v^{1/2-s}\,dv
$$
This converges if $\Re(s) > \frac{3}{2}$.  In that case,
$$
    I = -\frac{π^2}{\frac{3}{2} - s} = \frac{2π^2}{2s-3}
$$
