Convergence of the Integral $\int d^d k \frac{1}{k^2}\exp\left({i\vec k \cdot \vec r}\right)$. The integral: 
 $$I(\vec r)=\int_{|\vec k|\lt \Lambda} d^d k \; \;\frac{1}{k^2}\exp\left({i\vec k \cdot \vec r}\right)\tag{1}$$
appears quite a lot in physics. I have seen the convergence of this studied by setting $\vec r =\vec 0$ and showing that:
$$I(\vec 0)=SA\int_{k\lt \Lambda} d k \;\;\frac{k^{d-1}}{k^2}\tag{2}$$
where $SA$ is the surface area of a hypersphere in $d$ dimensions. This is clearly (IR) divergent for $d\le 2$. My question is:
Is the divergence/convergence of (1) totally determined by that of (2) and if so why?
 A: Since I am a physicists rather then a mathematician this answer is not going to be  very mathematically rigorous.
Convergence of (2) $\Rightarrow$ Convergence of (1)
We have:
$$(1)=\int_{|\vec k|\lt \Lambda} d^d k \; \;\frac{1}{k^2}\exp\left({i\vec k \cdot \vec r}\right)$$
$$\left|(1)\right|=\left|\int_{|\vec k|\lt \Lambda} d^d k \; \;\frac{1}{k^2}\exp\left({i\vec k \cdot \vec r}\right)\right|$$
$$\le \int_{|\vec k|\lt \Lambda} d^d k \; \;\left|\frac{1}{k^2}\exp\left({i\vec k \cdot \vec r}\right)\right|\tag{Triangle inequality}$$
$$=\int_{|\vec k|\lt \Lambda} d^d k \; \;\frac{1}{k^2}=(2)$$
Thus
$$\left|(1)\right|\le (2)$$
so if (2) converges (1) must converge.
Divergence  of(2) $\Rightarrow$ Divergence of (1)
This is slightly more tricky I am going to write $d^dk=k^{d-1} dk d\Omega$ for the 'solid angle' $\Omega$. Then:
$$(1) =\int d\Omega \int^\Lambda_0 dk \frac{1}{k^2}k^{d-1} \exp(i\vec k\cdot \vec r)$$
$$=\int d\Omega\left(\int^\Lambda_\varepsilon \frac{1}{k^2}k^{d-1} \exp(i\vec k\cdot\vec r)+\int^\varepsilon_0 \frac{1}{k^2} k^{d-1}\exp(i\vec k\cdot \vec r)\right)$$
The first term is convergent. For the second term we have that:
$$\int d\Omega\int^\varepsilon_0 \frac{1}{k^2} \exp(i\vec k\cdot \vec r)=\int d\Omega\int^\varepsilon_0 \frac{1}{k^2} k^{d-1}\cos(\vec k \cdot \vec r)+i\int d\Omega\int^\varepsilon_0 \frac{1}{k^2}k^{d-1} \sin(\vec k \cdot \vec r)$$
Thus
$$\left|\int^\varepsilon_0 \frac{1}{k^2} k^{d-1}\exp(i\vec k\cdot \vec r)\right|\ge \cos(\varepsilon r\cos(\theta))\int_0^\varepsilon k^{d-1}\frac{1}{k^2} dk$$
Since  we can define $\varepsilon$ sufficiently small such that
$$\cos(\varepsilon r\cos(\theta))\ge C$$
for some $C\ne 0$ and arbitary:
$$\left|\int^\varepsilon_0 \frac{1}{k^2} \exp(i\vec k\cdot \vec r)\right|\ge C\int_0^\varepsilon\frac{1}{k^2} k^{d-1}dk$$
If
$$\int_0^\varepsilon\frac{1}{k^2} k^{d-1}dk$$
diverges then so will the $\int^\varepsilon_0k^{d-1} \frac{1}{k^2} \exp(i\vec k\cdot \vec r)$ which is that that we wanted to prove.
