$$\int_{spacetime}\frac{d^4x}{(x^2)^2}=\int_{-\infty}^{\infty} dt\int_{-\infty}^{\infty} dx\int_{-\infty}^{\infty} dy\int_{-\infty}^{\infty} dz\frac{1}{(t^2-x^2-y^2-z^2)^2}$$ To show that $\int_{spacetime}\frac{d^4x}{(x^2)^2}$ diverges in physics we use this type of non rigourous arguments $$d^4x\approx k|x|^3d|x|\implies \int_{spacetime}\frac{d^4x}{(x^2)^2}=\int_{0}^{\infty}\frac{k|x|^3d|x|}{|x|^4}=\int_{0}^{\infty}{kd(ln|x|)}=\text{diverges}$$
Can someone rigorously prove the above relation? Example here they did like that.
The main problem is $x^2$ can be $<0$ so $|x|$ can be imaginary. So the above method is not straightforward without proper justification.
You can always use Tonelli's theorem and then go to the usual 3D polar coordinates. SO, the integral $I$ can be calculated as \begin{align} I&=\int_{\Bbb{R}^3}\int_{\Bbb{R}}\frac{1}{(t^2-x^2-y^2-z^2)^2}\,dt\, d(x,y,z)\tag{Fubini-Tonelli}\\ &=\int_{0}^{\infty}\int_{\Bbb{R}}\frac{1}{(t^2-r^2)^2}\,dt\cdot 4\pi r^2\,dr\tag{Polar coordinates}\\ &=8\pi\int_0^{\infty}\int_0^{\infty}\frac{r^2}{(t^2-r^2)^2}\,dt\,dr\tag{evenness} \end{align} Now, for each $r\in (0,\infty)$, we have $\int_0^{\infty}\frac{r^2}{(t^2-r^2)^2}\,dt=r^2\int_0^{\infty}\frac{1}{(t-r)^2(t+r)^2}\,dt=\infty$, because the singularity at $t=r$ is a quadratic $\frac{1}{(t-r)^2}$, and these are not integrable singularities (look up the integral p-tests from single variable calculus). Since for each $r\in (0,\infty)$ the inner integral is $\infty$, the whole thing is $\infty$: \begin{align} I&=8\pi\int_0^{\infty}\infty\,dr=\infty. \end{align}