How to compute $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{[-1,1]^n}\exp[2\pi i(\theta_1 v_1.x+\theta_2v_2.x)]d^nx d\theta_1d\theta_2$

Question

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 $e(z) = e^{2 \pi i z}$

Answer 1

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 $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla g(x)|}\,d^{n-1}\sigma(x),$$ 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.}