Let $F(\mathbf{x}) \in \mathbb{Z}[x\_1, ..., x\_n]$ be a degree $d$ homogeneous form. Let
$$
I(\alpha) = \int\_{[0,1]^n} e^{2 \pi i F(\mathbf{x}) \alpha} dx\_1...dx\_n.
$$
Then the singular integral is defined as 
$$
\sigma\_{\infty} = \int\_{\mathbb{R}} I(\alpha) d\alpha.
$$


It says in an article I am reading that $\sigma\_{\infty} > 0$
if the equation $F(\mathbf{x}) =0$ has a non-singular real solution in $(0,1)^n$.
I would greatly appreciate if someone could give me a hint or some explanation on how I can prove this statement. Thank you very much.



Showing the positivity of a singular integral that came up in circle method
Here's a sketch. First of all, proceeding formally, if we start with the $\alpha$ integral, then we obtain
$$
\sigma\_{\infty} = \int \delta(F(x))\, dx ,
$$
and near an $a$ with $F(a)=0$, $\nabla F(a)\not= 0$, this looks like $\delta(v\cdot t)$, so we obtain a positive contribution.


To make a proof out of this, fix a $\varphi\ge 0$ with $\widehat{\varphi}\ge 0$, $\varphi(0)=1$ (Fejer kernel would work), and consider
$$
\int I(\alpha)\varphi(\alpha/n)\, d\alpha = n \int \widehat{\varphi}(nF(x))\, dx .
$$
Now near an $a=(b,c)$ as above, with $b\in\mathbb R$, $c\in\mathbb R^{d-1}$, we find solutions to $F=0$ in a whole neighborhood $U\subseteq \mathbb R^{d-1}$ of $c$ (if $\partial F/\partial x\_1(a)\not=0$), by the implicit function theorem. As before, integration over $(a-\delta,a+\delta)\times U$ gives a contribution that stays positive in the limit $n\to\infty$.


