To begin with, it's known that $$V(q)=\int d^{2} r e^{-i q \cdot r} \frac{1}{r}=\frac{2\pi}{q}$$ but I would like to understand as much of this calculation as possible.
We can begin calculating the Fourier transform of the 2D Coulomb potential with $$V(q)=\int d^{2} r e^{-i q \cdot r} \frac{1}{r}=\int_{0}^{\infty} r d r \int_{0}^{2 \pi} d \theta e^{-i q r \cos \theta} \frac{1}{r}=\int_{0}^{\infty} d r \int_{0}^{2 \pi} d \theta e^{-i q r \cos \theta}$$ Then we can use one of Bessel's integrals $$2 \pi J_{0}(x)=\int_{0}^{2 \pi} d \theta e^{-i x \cos \theta}$$ to obtain $$V(q)=\int_{0}^{\infty} d r \int_{0}^{2 \pi} d \theta e^{-i q r \cos \theta}=2 \pi \int_{0}^{\infty} d r J_{0}(q r)=\frac{2 \pi}{q} \int_{0}^{\infty} q d r J_{0}(q r)=\frac{2 \pi}{q} \int_{0}^{\infty} d u J_{0}(u)$$ Apparently the integral $$\int_{0}^{\infty} d u J_{0}(u)=1$$ but I have not found a proof of this.
Where can I find a proof of this? I have found proofs of Bessel's integrals but not this definite integral.
We can use the Laplace transform of the Bessel function, which is given by (see here) $$ L[J_0(z)] = \int_0^\infty J_0(z) e^{-st}dz =\frac{1}{\sqrt{s^2 +1}} $$ Now insert $s=0$. Viola.