I would like to integrate $$ \int d^3 \vec{k} \int d^3 \vec{q} \frac{1}{q^2} $$ under the condition that $$ |\vec{k}| \leq A\ \ \ \text{and}\ \ \ |\vec{k} + \vec{q}| \leq A$$ where $A$ is some constant.
Is this executable? I have no idea how to handle (angle term of) the second condition.
On
The integral wrt $\mathbf k$ is the volume of the intersection of two balls centered at the origin and at $-\mathbf q$. The formula for the volume of a solid of revolution gives $$V = 2 \pi \int_{q/2}^A (A^2 - x^2) dx \, [q < 2 A] = \frac \pi {12} (q + 4 A) (q - 2 A)^2 \, [q < 2 A].$$ Then the original integral is $$2 \pi \int_0^{2 A} \int_0^\pi V \sin \theta \, d\theta dq = 4 \pi^2 A^4.$$
This integral is in fact analytically tractable.
Denote $|\mathbf{k}|=k$ for any vector and first consider a change of variables $\mathbf{q+k}=R\mathbf{p}$, $R^TR=1$ such that $R^T\mathbf{k}=k\mathbf{\hat{z}}$ and then regular spherical coordinates. Then we obtain
$$\int_{|\mathbf{q}+\mathbf{k}|\leq A} d^3q\frac{1}{q^2}=\int_{p\leq A} d^3p\frac{1}{(\mathbf{p}-k\mathbf{\hat{z}})^2}=2\pi\int_0^Aq^2dq\int_{-1}^1d(\cos\theta)\frac{1}{q^2+k^2-2qk\cos\theta}\\=\frac{2\pi}{k}\int_0^Aqdq\ln\Big|\frac{q+k}{q-k}\Big|=\frac{\pi A^2}{k}\big(\ln\frac{A+k}{A-k}+1+\frac{k^2}{A^2}\ln\frac{A^2-k^2}{k^2}\Big)$$
This can be integrated once again without problem in spherical coordinates,since it depends only on the magnitude of the vector $\mathbf{k}$:
$$\int_{k\leq A}d^3k \frac{\pi A^2}{k}\big(\ln\frac{A+k}{A-k}+1+\frac{k^2}{A^2}\ln\frac{A^2-k^2}{k^2}\Big)\\=4\pi^2 A^4\int_{0}^1 xdx\Big(\ln\frac{1+x}{1-x}+1+x^2\ln\frac{1-x^2}{x^2}\Big)=4\pi^2A^4(\frac{5}{6}-\frac{2}{3}\ln2)$$
I haven't checked these integrals against Alpha, so I am accepting corrections :)