Here is yet another solution, which probably is not a physicist's way of thinking.
After <PERSON> rotation, we may work on Euclidean space. The function $f(k_E) = |k_E|^{-4}$ is not square-integrable on $\mathbb{R}^4$, however, so its Fourier transform does not exist in ordinary sense. A moment of thought suggests that it can be realized as distribution on the space
$$ \mathcal{A} := { \varphi \in \mathcal{S}(\mathbb{R}^4) : \textstyle \int_{\mathbb{R}^4} \varphi(\epsilon) \, \mathrm{d}^4\epsilon = 0 }, $$
where $\mathcal{S}(\mathbb{R}^4)$ is the <PERSON> space. 578
Then computing
$$I(\epsilon) = i \check{f}(\epsilon) = \frac{i}{(2\pi)^4} \int_{\mathbb{R}^4} |k_E|^{-4} e^{i\epsilon \cdot k_E} \, \mathrm{d}^4 k_E$$
in distribution sense amounts to identifying the following pairing
$$ \langle I, \varphi \rangle = \langle \check{f}, \varphi \rangle = \langle f, \check{\varphi} \rangle, \qquad \forall \varphi \in \mathcal{A}. $$
Since $\check{\varphi}(0) = 0$ and $\check{\varphi}$ has rapid decay near infinity, the pairing $\langle f, \check{\varphi} \rangle$ is realized as <PERSON> integral. Then by the Fubini's theorem,
\begin{align} \langle f, \check{\varphi} \rangle &= i \int_{\mathbb{R}^4} \frac{1}{|k_E|^4} \check{\varphi}(k_E) \, \mathrm{d}^4 k_E = i \int_{\mathbb{R}^4} \bigg( \int_{0}^{\infty} t \mathrm{e}^{-|k_E|^2 t} \, \mathrm{d}t \bigg) \check{\varphi}(k_E) \, \mathrm{d}^4 k_E, \ &= i \int_{0}^{\infty} t \bigg( \int_{\mathbb{R}^4} \mathrm{e}^{-t |k_E|^2} \check{\varphi}(k_E) \, \mathrm{d}^4 k_E \bigg) \mathrm{d}t. 592