In the appendix A of this paper by Braaten et al., the authors try to compute the divergences of two integrals that come from an expansion of an action $I$ in $\langle e^{iI} \rangle$, via dimensional regularization of the propagator, given by
$$\tag1 D^{(1,1)} = i (2\lambda^2)^{-1} \int \text{d}^2x B_\mu^{ki}(x) B^{\mu ki}(x) \langle \zeta^i(x) \zeta^j(x)\rangle$$
$$ \tag2 D^{(1,2)} = \frac{i^2}{2} (2\lambda^2)^{-2} \int \text{d}^2x\int \text{d}^2y B_\mu^{ij}(x) B_\nu^{ kl}(y) \ 4\langle (\zeta^i \partial^\mu \zeta^j)(x)(\zeta^k\partial^\nu \zeta^l)(y)\rangle.$$
Here, indices $i,j,k,...$ are spacetime indices and $\mu, \nu$ are indices that run from $1,2$, $B_\mu^{ij}$ are gauge fields antisymmetric in the latin indices, and the feynman propagator of the theory is given by (in the limit $m \to 0^+$)
$$\tag3 \langle \zeta^i(x) \zeta^j(y)\rangle = \frac{i\lambda^2 \delta^{ij}}{(2\pi^2)} \int \ \frac{e^{i p \cdot (x-y)}}{p^2-m^2} \text{d}^{d=2+\epsilon}p.$$
Right bellow $(1)$ and $(2)$ it is said
Each of these diagrams is ultraviolet divergent. However, their sum is finite, for arbitrary background potentials $B_\mu$, corresponding to the well-known ultraviolet finiteness of the self-energy for a minimally coupled gauge field in two dimensional spacetime. This may be seen explicitly to one-loop using the integral $$ I_{\mu \nu} = (2\pi)^{-2} \int \text{d}^d p \frac{p_\mu p_\nu}{(p^2 -m^2)^2}= \frac{i}{4\pi}\delta_{\mu \nu} \frac{1}{d-2} + \cdots, \tag{A.37}$$which arises upon evaluating the ultraviolet divergent part of the expectation value appearing in $D^{(1,2)}$. It follows that the ultraviolet divergence in $D^{(1,2)}$ cancels that in $D^{(1,1)}$.
Evaluating the $D^{(1,1)}$ we get
\begin{align} \tag4 D^{(1,1)} &= i (2\lambda^2)^{-1} \int \text{d}^2x B_\mu^{ki}(x) B^{\mu ki}(x) \frac{i \lambda^2}{(2\pi)^2}\int \text{d}^dp \frac{1}{p^2-m^2}\\ &=- \frac{1}{2}\int \text{d}^2x B_\mu^{ki}(x) B^{\mu ki}(x)\left( \frac{i}{2\pi} \frac{1}{d-2} + \cdots \right)\\ &= -\frac{i}{4\pi} \frac{1}{d-2} \int \text{d}^2xB_\mu^{ki}(x) B^{\mu ki}(x) + \cdots. \end{align}
However, what I have done so far with $D^{(1,2)}$ was
\begin{align} \tag5 D^{(1,2)} &= \frac{i^2}{2} (2\lambda^2)^{-2} \int \text{d}^2x\int \text{d}^2y B_\mu^{ij}(x) B_\nu^{ kl}(y) \ 4\langle (\zeta^i \partial^\mu \zeta^j)(x)(\zeta^k\partial^\nu \zeta^l)(y)\rangle\\ &= \frac{1}{2} \frac{1}{(2\pi^4)} \int \text{d}^2x\int \text{d}^2y B_\mu^{ij}(x) B_\nu^{ ij}(y) \ \int \text{d}^dp \text{d}^dk \frac{e^{i(p+k)(x-y)} k^\mu (k^\nu - p^\nu)}{(p^2 -m^2)(k^2-m^2)}, \end{align}
but I don't know how to proceed. By comparing $(4)$ with $(5)$, using $(\text{A},37)$, I should obtain two delta functions $\delta(p+k)$ and $\delta(x-y)$, but I cannot use the inverse fourier form of the delta in the above integral directly. I will aprecciate any help
