Let $F\_1(\mathbf{x}), F\_2(\mathbf{x}) \in \mathbb{Z}[x\_1, ..., x\_n]$ be a degree $d$ homogeneous polynomial. 
For the system of equations $$F\_1(\mathbf{x})= F\_2(\mathbf{x}) =0,$$
we have the following singular integral
$$
\sigma\_{\infty} = \int\_{\mathbb{R}^2} I(\alpha\_1, \alpha\_2) d\alpha\_1d\alpha\_2,
$$
where
$$
I(\alpha\_1, \alpha\_2) = \int\_{[0,1]^n} e^{2 \pi i (F\_1(\mathbf{x}) \alpha\_1 + F\_2(\mathbf{x}) \alpha\_2 )
} dx\_1...dx\_n.
$$


Let $G\_1(\mathbf{x}) = F\_1(\mathbf{x}) + g(\mathbf{x})F\_2(\mathbf{x})$
and $G\_2(\mathbf{x}) = F\_2(\mathbf{x})$, where $g$ is some homogeneous polynomial. Then the set of solutions of 
$$G\_1(\mathbf{x})= G\_2(\mathbf{x}) =0$$
is the same as that of $F\_1$ and $F\_2$.
Let $\sigma'\_{\infty}$ be the singular integral of this system obtained by replacing $F\_1$ and $F\_2$ with $G\_1$ and $G\_2$, respectively, in the definition of singular integral above. 


I suspect that $\sigma\_{\infty}$ and $\sigma'\_{\infty}$ are the same, since they both correspond to the same affine variety, but I don't see how I can prove this statement. Could anyone please give me a hint or explanation on how I can see this (assuming the integrals exists, etc)? Thank you very much.



Is the singular integral that come up in circle method independnet of the representatin of the equations?
Formally (ignoring issues of integrability), one has (writing $e(\theta) := e^{2\pi i\theta}$)


$$ \sigma'\_\infty = \int\_{{\bf R}^2} \int\_{[0,1]^n} e( F\_1({\bf x}) \alpha\_1 + g({\bf x}) F\_2({\bf x}) \alpha\_1 + F\_2({\bf x}) \alpha\_2)\ d{\bf x} d\alpha\_1 d\alpha\_2$$
$$ = \int\_{{\bf R}} \int\_{[0,1]^n} e( F\_1({\bf x}) \alpha\_1) ( \int\_{\bf R} e( (\alpha\_2 + g({\bf x}) \alpha\_1) F\_2({\bf x}))\ d\alpha\_2 ) d{\bf x} d\alpha\_1.$$
Similarly for $\sigma\_\infty$ without the $g({\bf x})$ term. If one formally translates $\alpha\_2$ by $g({\bf x}) \alpha\_1$ in the inner integral, one obtains the desired identity $\sigma\_\infty = \sigma'\_\infty$.


To make this rigorous, one has to damp the ${\bf R}^2$ integral by some mollifier (e.g. a gaussian $e^{-\varepsilon\_1 \alpha\_1^2 - \varepsilon\_2 \alpha\_2^2}$, and it will be convenient to make $\varepsilon\_2$ a bit smaller than $\varepsilon\_1$ so that the aforementioned translation has negligible impact on the mollifier) in order to justify the various changes of variable, and use suitable decay bounds on oscillatory integrals (related to whatever bounds were needed to ensure the absolute integrability of $I$) to take limits properly.


From the point of view of the theory of distributions, one morally has
$$ \sigma\_\infty = \int\_{[0,1]^n} \delta(F\_1({\bf x})) \delta(F\_2({\bf x}))\ d{\bf x}$$
where $\delta$ is the Dirac delta, which makes the claimed invariance formally obvious. If the variety $\Sigma := \{ {\bf x} \in [0,1]^n: F\_1({\bf x}) = F\_2({\bf x}) = 0 \}$ is smooth, this expression can also be written as
$$ \int\_{\Sigma} \frac{dS}{|dF\_1 \wedge dF\_2|},$$
where $dS$ is surface measure, which is another way to make the claimed invariance manifest.


