Converting between $\frac{d \sigma}{d t}$ and $\frac{d \sigma}{d \Omega}$ I'm performing some simulation where I require the integration of a  cross section of the form 
$$\sigma = \int_{-s}^{0}dt \frac{1}{(t-\mu^2)^2} f(\theta,\phi)$$
where $s$ and $t$ are the usual mandelstam variables, $\mu^2$ is some constant, and  $f$ is some non-Lorenz invariant function of the background and scattering angles. Because of this $f$ I cannot boost to the COM frame where the kinematic become trivial, therefore I want to rather integrate 
$$ \sigma = \int d\theta d\phi \frac{1}{(t(\theta,\phi)-\mu^2)^2} f(\theta,\phi) $$
what I'm finding non-trivial is determining the Jacobian of this transformation (since we're going from a 1-D to a 2-D integral the Jacobian matrix is non-square). 
The big problem is that $t(\theta,\phi)$ is a degenerate function of $\theta$ and $\phi$ and therefore the integration  over $\theta$ and $\phi$ double counts my differential cross section and therefore the total cross section I end up with is too large.
I checked this via numerical integration for $f=1$ and compared to the analytic result
$$\sigma = \int_{-s}^{0}dt \frac{1}{(t-\mu^2)^2} = \frac{s}{\mu^2(\mu^2+s)} $$
I tried a kind of brute force method to find the jacobian, by extracting lines of constant $t$ from a Mathematica contour plot, and calculating the arclength of these lines, and then divide out the degeneracy.
Long story short, it doesn't work.
tl;dr 
Find the jacobian which converts $$ \int_{-s}^{0} dt \rightarrow \int d\theta d\phi$$ in the lab frame
Any ideas?
EDIT:
So my angles are defined in the following way, I align my z-axis with one of the incoming particles, specifying the $\theta$ and $\phi$ (i.e the unit vector) into which it scatters uniquely determines its magnitude and then trivially (by momentum conservation) the other scattering vector. (since we are not in COM frame the other incoming vector isn't anti-aligned)
EDIT 2: In an attempt to make it well-posed I can make the following "change"
$$
\int_{-s}^{0} dt = \int_{0}^{2\pi}\int_{-s}^{0} d\psi dt \frac{1}{2\pi} $$
In the COM frame this would correspond to integrating over the cone of vectors that correspond to a specific value of $t$. In principle I should be able to crunch out a Jacobian now, thats if I can express $\psi(\theta,\phi)$ ( which would correspond to lines of constant $t$).....hmmmmm....I'm going to run with this idea for a bit. 
 A: Independent of the physical process, this question is mathematically ill-posed. If you want to convert the integral from the 1D form that you gave to the 2D form that you gave, you are not making an "ordinary" change of variables where you can use a Jacobian matrix to adjust the measure. (As you noted yourself, the Jacobian in this case won't be square, so operationally you won't be able to apply this formula even before you get into the details of whether it's the correct formula to use.)
If you really want to make this change, then you need to be able to write one or the other of the variables as a function of the other, i.e. $\theta = \theta(\phi)$ or $\phi = \phi(\theta)$.  Then you can write your integral as two nested, 1D integrals.  In general, however, the limits for the integration on the inner integral will be a function of the variable of integration in the outer integral.  In symbols, you'll end up with something like this:
$$ \sigma = \int_a^b \left[ \int_{l(\theta')}^{u(\theta')} \frac{f(\theta(\phi'),\phi')}{(t(\theta(\phi'),\phi')-\mu^2)^2} \delta(\theta(\phi') - \theta') \frac{dt}{d\phi'}  d\phi' \right] d\theta' $$.
The $\delta$ is the Dirac $\delta$-function.
Moreover, since you said that $t$ is highly degenerate as a function of the angles, you'll actually need to break the integral above into multiple domains such that $t$ is not degenerate over the range of integration of each subdomain.  You then sum the results of the integrals on the subdomains.
Unless your problem has some additional structure that you can exploit, making these changes will probably be as hard or harder than evaluating the original integral. If you cannot evaluate it analytically, then you might want to move to a numerical integration technique or find some other transformation for the integral that applies to your problem.
If you insist on a "simple" answer to your question, I guess you would say that $\delta(\theta(\phi') - \theta') \frac{dt}{d\phi'}$ is the determinant of the Jacobian that you sought, although that washes out the points about the limits of integration being non-trivial and the possible impacts of the degeneracy of $t$ as a function of the angles.
