Let $\mathbf{v}\_1, \mathbf{v}\_2$ be two vectors in $\mathbb{R}^n$. I would like to compute the following singular integral:
$$\int\_{-\infty}^{ \infty} \int\_{-\infty}^{\infty} 
\int\_{[-1,1]^n}
e(\theta\_1 \mathbf{v}\_1.\mathbf{x} +\theta\_2 \mathbf{v}\_2.\mathbf{x} ) d \mathbf{x} d \theta\_1 d\theta\_2$$


I vaguely understand that this is a volume of the subset of $[-1, 1]^n$ where
$$\mathbf{v}\_1.\mathbf{x} = \mathbf{v}\_2.\mathbf{x} = 0$$ and $.$ denotes the dot product. But I'm struggling to show this rigorously... Any reference/suggestions is appreciated!


We use the notation $e(z) = e^{2 \pi i z}$



How to compute $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{[-1,1]^n}\exp[2\pi i(\theta_1 v_1.x+\theta_2v_2.x)]d^nx d\theta_1d\theta_2$
First integrate over $\theta\_1,\theta\_2$. Use the delta function representation (for $k\in\mathbb{R}$)
$$\int\_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$
to evaluate
$$\int\_{-\infty}^\infty \int\_{-\infty}^\infty e^{2\pi i (v\_1\cdot x)\theta\_1+2\pi i(v\_2\cdot x)\theta\_2}\,d\theta\_1 d\theta\_2=\delta(v\_1\cdot x)\delta(v\_2\cdot x).$$


Next for the integral over the vector $x\in\mathbb{R}^n$ use the [coarea formula](https://en.wikipedia.org/wiki/Dirac_delta_function#Properties_in_n_dimensions)
$$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int\_{g(x)=0}\frac{f(x)}{|\nabla g(x)|}\,d^{n-1}\sigma(x),$$
once with $g(x)=v\_1\cdot x$, $f(x)=\delta(v\_2\cdot x)$, then once more with $g(x)=v\_2\cdot x$, $f(x)=1$, to arrive at
$$\int\_{[-1,1]^n}\delta(v\_1\cdot x)\delta(v\_2\cdot x)\,d^nx=\frac{1}{|v\_1||v\_2|}\int\_{[-1,1]^n\cap v\_1\cdot x=0\cap v\_2\cdot x=0}d^{n-2}\sigma(x).$$


So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v\_1\cdot x=0=v\_2\cdot x$, divided by the product $|v\_1||v\_2|$ of the lengths of the two vectors $v\_1,v\_2$.


Dimensional check: if $x$ has dimension of length $L$ and $\theta\_1,\theta\_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v\_1||v\_2|$ has dimension of $L^{-2}$ the dimensions match.


