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$$\int_{spacetime}\frac{d^4x}{(x^2)^2}=\int_{-\infty}^{\infty} dt\int_{-\infty}^{\infty} dx\int_{-\infty}^{\infty} dy\int_{-\infty}^{\infty} dz\frac{1}{(t^2-x^2-y^2-z^2)^2}$$ 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.

Kasi Reddy Sreeman Reddy
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  • The fact that you're considering the metric with Lorentz signature makes no difference at all here. This is simply a plain old vanilla integral in $\Bbb{R}^4$. – peek-a-boo Mar 25 '22 at 19:33
  • @peek-a-boo how is it the same? The magnitude of $x$, that is $|x|$ can be a complex number. – Kasi Reddy Sreeman Reddy Mar 25 '22 at 19:34
  • That's just a bit of (abuse of) notation. The definition of the integral still remains the same, so forget about the fact that you have Lorentz signature, and figure out if $\int_{\Bbb{R}^4}\frac{1}{(t^2-x^2-y^2-z^2)},dx,dy,dz,dt$ is finite or not. One can easily solve the problem (as always even in basic multivariable calculus) using Fubini/Tonelli's theorem. – peek-a-boo Mar 25 '22 at 19:36
  • @peek-a-boo I don't know how to do that integral in $\Bbb{R}^4$ that's why we go to polar like coordinates in Lorentz metric to do it. – Kasi Reddy Sreeman Reddy Mar 25 '22 at 19:38

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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}

peek-a-boo
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  • These uses of $\infty$ are completely legitimate; it is completely standard to deal with non-negative, even infinite valued functions when dealing with standard Lebesgue integrals. If you want to confine yourself to Riemann integrals, then you have to consider everything as an improper integral, and take appropriate limits. THe net result is still the same: the final integral is still $\infty$. – peek-a-boo Mar 25 '22 at 19:48
  • Thanks, I understood your method to get the answer in the 3d cartesian+time separate system. But can you also answer whether the method using Lorentz metric space is correct or wrong? – Kasi Reddy Sreeman Reddy Mar 25 '22 at 19:52
  • @KasiReddySreemanReddy frankly I stopped reading that approach when I saw the use of $|x|=\sqrt{t^2-x^2-y^2-z^2}$, which is complex valued (and you didn't even specify a branch for the square root to make things well-defined). Likewise, without further explanation $d|x|$ is meaningless. So honestly, it takes way too much effort to make that line of reasoning even make a slight bit of sense. You asked for a rigorous proof, and I gave you one (which frankly I think is infinitely simpler to come up with and understand) – peek-a-boo Mar 25 '22 at 19:55
  • Okay that is true, that method is far too vague. Ok then your answer is sufficient for me. – Kasi Reddy Sreeman Reddy Mar 25 '22 at 19:57