During a calculation of Feynman diagram, I encountered an integral which is diverging: $$\int d^{2}p\frac{p^{2}p_{n}}{\left(p^{2}+\alpha k^{2}\right)^{2}}e^{ip\cdot y}$$ where $p$, $k$ and $y$ are vectors of two components, and $n$ can be either 1 or 2.
In order to perform the integration, the only thing I managed to think of, is to expand the exponent and then integrate term by term after using dimensional regularization in order to cure the divergence.
Unfortunately, each term in this series is diverging and therefore I'm not sure that integrating term-by-term (using $\int f+g=\int f+\int g$) is allowable in this case.
I would like to ask two questions: A. How can I justify integration term-by-term (if it is allowable)?
B. Is there any other way to do the integration?
In order to make it more transparent, the integral above can be written in component form also as: $$\int dp_{1}\, dp_{2}\frac{\left(p_{1}^{2}+p_{2}^{2}\right)p_{n}}{\left(p_{1}^{2}+p_{2}^{2}+\alpha(k_{1}^{2}+k_{2}^{2})\right)^{2}}e^{ip_{1}y_{1}+ip_{2}y_{2}}$$
It is probably easier to use polar coordinate system $p=(p_1,p_2)=p(\cos \theta,\sin\theta),y=(y_1,y_2)=y(\cos \phi,\sin\phi)$:
$$\int d^{2}p\frac{p^{2}p_{1}}{\left(p^{2}+\alpha k^{2}\right)^{2}}e^{ip\cdot y}= \int_0^{2\pi}d\theta\int_0^{\infty} pdp\frac{p^{2}p \cos\theta}{\left(p^{2}+\alpha k^{2}\right)^{2}}e^{ip y \cos(\theta-\phi)}$$ $$=2i\pi \cos(\phi)\int_0^{\infty} pdp\frac{p^{2}p J_1(py)}{\left(p^{2}+\alpha k^{2}\right)^{2}}$$
$$\int d^{2}p\frac{p^{2}p_{2}}{\left(p^{2}+\alpha k^{2}\right)^{2}}e^{ip\cdot y}= \int_0^{2\pi}d\theta\int_0^{\infty} pdp\frac{p^{2}p \sin\theta}{\left(p^{2}+\alpha k^{2}\right)^{2}}e^{ip y \cos(\theta-\phi)}$$ $$=2i\pi \sin(\phi)\int_0^{\infty} pdp\frac{p^{2}p J_1(py)}{\left(p^{2}+\alpha k^{2}\right)^{2}}$$