How Does Renormalization Address Divergences in Integrals?

[eljose]

for a divergent series i can write an expression in the form:

$$\int_{R}dxC(x)w(x)e^{-ax}$$

where a is a divegent quantity in the form $$a=ln\epsilon$$

the qeustion is how i would apply renormalization?..in fact if we apply functional differentiation respect to e^{-ax} we get

$$C(x)w(x)$$ the divergent term magically disappears

 

[quantumdude]

Moving from Homework Help.

 

[tacman]

Renormalization is a powerful tool used in theoretical physics to handle divergent series and integrals. It is a mathematical technique that allows us to remove the infinities and still obtain meaningful results. In the context of your question, renormalization can help us deal with the divergent quantity "a=ln\epsilon" in the integral expression.

To apply renormalization in this case, we first need to understand the concept of a regularization parameter. In this case, "a=ln\epsilon" is the regularization parameter which is used to regulate the divergence in the integral. The idea behind renormalization is to replace this regularization parameter with a physical parameter that has a finite value.

To do this, we can first rewrite the integral expression as:

\int_{R}dxC(x)w(x)e^{-ax} = \int_{R}dxC(x)w(x)e^{-a_{R}x + a_{R}x - ax}

Here, a_{R} is a new parameter that we have introduced. Now, we can expand the exponential term using the Taylor series and keep only the first two terms:

e^{-a_{R}x + a_{R}x - ax} \approx 1 - (a_{R} - a)x + \mathcal{O}(x^2)

Substituting this back into the integral expression, we get:

\int_{R}dxC(x)w(x)(1 - (a_{R} - a)x + \mathcal{O}(x^2))

Now, we can see that the integral contains a term that is linear in x, which will give rise to the divergence. To remove this divergence, we can use the method of dimensional regularization. This involves replacing the variable x with a dimensionless quantity x/\mu, where \mu is a finite parameter. This effectively rescales the integral and removes the divergence.

After performing this rescaling, we are left with an integral that is finite and can be evaluated. This is the renormalized integral and it is free of any divergences. We can now set \mu to a physical value and obtain meaningful results.

In summary, renormalization is a powerful technique that allows us to remove infinities from our calculations and obtain physically meaningful results. By introducing a new parameter and using dimensional regularization, we can rescale the integral and remove the divergence, resulting in a finite and well-defined integral.