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I'm studying regularization of divergent integrals in QFT from Here: Roberto Soldati - Field Theory 2. Intermediate Quantum Field Theory (A Next-to-Basic Course for Primary Education)

I think I'm missing the big picture, let me explain.

At page 166 he says

In order to give a precise mathematical meaning to the above listed ill defined integral expressions, we have to introduce from the outset some kind of regularization procedure, the aim of which is to build up absolutely convergent loop integrals.

Bold mine.

Then pages later (from page 178), when he explains dimensional regularization he does it with a paradigmatic example, he considers this integral:

$$ I = \mu^{4 - 2\omega} \int \displaystyle\frac{d^{2\omega} k_E}{(2\pi)^{2\omega}} \big(k_E^2 - \Delta \big)^{-2} $$

We then obtain (page 181 equation 4.23)

$$ I = \frac{1}{16 \pi^2} \bigg( \displaystyle\frac{1}{\epsilon} - {\bf{C}} + ln \displaystyle\frac{4\pi\mu^2}{\Delta} \bigg) + O(\epsilon) \tag{4.23} $$

Where $\epsilon = 2-\omega$.

Now if we want $I$ in 4 dimensions we have to take $\epsilon = 0$ and we see in $(4.23)$ that with this choice $I$ diverges because of the term $\displaystyle\frac{1}{\epsilon}$, therefore we didn't built a convergent integral as stated in the bolded part of the quote above. Then I ask:

What did we achieve with this procedure if when we come back to four dimension we still have a divergent result?

And since I am probably missing the point and the big picture please feel free to explain it to me.

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    $\begingroup$ You may want to know the next step,i.e. renormalization scheme, after the step of regularization in the QFT renormalization program, when one will eventually compute something observable like scattering amplitude. physics.stackexchange.com/questions/495919/… $\endgroup$ – chichi Aug 14 '19 at 5:59
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The integral is absolutely convergent for $\epsilon >0$. The point is the following. Frequently in QFT we care about an amplitude $A$ which can be expressed as the sum of multiple integrals

$$A = I_1 + I_2 + \cdots + I_n.$$

Each integral, roughly, corresponds to a different Feynman diagram. The sum $A$ is the only physical quantity we care about, and therefore it must be finite. For reasons having to do with both UV and IR divergences, each integral $I_i$ may be infinite, but in such a way that the infinities cancel out. This is the program of renormalization (for the UV divergences) and inclusive summation / KLN theorem (for the IR divergences).

So as you know we regularize each by (for instance) writing $d=4-\epsilon$, evaluating them all, summing them together, and then taking $\epsilon \to 0$ at the end. So indeed $\lim_{\epsilon\to 0} I_i$ may equal $\infty$ for any single integral, but if we have understood the physics correctly then $$\lim_{\epsilon\to 0} (I_1 + \cdots + I_n) = \text{finite.}$$

In other words the problematic $\frac{1}{\epsilon}$ (or higher power) terms cancel.

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    $\begingroup$ It is not strictly true that the sum converges in the $\varepsilon \rightarrow 0$ limit (unless maybe in a few simplified examples). The renormalized theory is recovered in the limit, but this isn’t the naive $\varepsilon \rightarrow 0$ limit, which is still divergent — the correct limit is $\varepsilon \rightarrow 0$, $g(\varepsilon) \rightarrow \infty$ with a certain form of the function $g(\varepsilon)$ which depends on the model. $\endgroup$ – Prof. Legolasov Aug 14 '19 at 10:37

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