Is Using Distribution Theory Overkill for Differentiating Under the Integral?

In summary, when evaluating the divergent integral of \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}}, it is important to consider the location of the poles and use the Cauchy Principal Value if they are on the real axis. If the poles are complex, the integral can be evaluated using the residue theorem. Additionally, differentiating with respect to a^{2} inside the integral can be justified using distribution theory, although other methods may work as well.
  • #1
zetafunction
391
0
would it be valid (in the sense of residue theorem ) the following evaluation of the divergent integral ?

[tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}}= \frac{ \pi i}{a} [/tex]

also could we differentiate with respect to [tex] a^{2} [/tex] inside the integral above to calculate

[tex] \int_{-\infty}^{\infty} \frac{dx}{(x^{2}-a^{2})^{2}} [/tex]
 
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  • #2
zetafunction said:
would it be valid (in the sense of residue theorem ) the following evaluation of the divergent integral ?

[tex] \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}}= \frac{ \pi i}{a} [/tex]

I believe you should consider two cases: the poles are on or off the real axis. If they are on the real-axis, then you can take the Cauchy Principal Value of the integral by considering a half-disc contour with indentations around the poles. The principal value in this case can be shown to be zero. If the poles are complex, then the integral over the same contour is equal to 2 pi i times the residue of the (single) enclosed pole with the integral over the half-circle arc as it's radius goes to infinity, going to zero and therefore, the real integral is 2 pi i times the residue of the enclosed pole.
 
  • #3
zetafunction said:
also could we differentiate with respect to [tex] a^{2} [/tex] inside the integral above to calculate

[tex] \int_{-\infty}^{\infty} \frac{dx}{(x^{2}-a^{2})^{2}} [/tex]

Yes it should be justifiable but then you would essentially be using real methods. I've seen a differentiation under the integral sign theorem in complex analysis but it seems useless since there exist much more general situations in which you can interchange differentiation and integration.
 
  • #4
You need to employ the notion of a principal value, otherwise the integral doesn't make sense - it depends on the way in which you take limits either side of the singularities. The best way to rigorously justify differentiating such things would be via the theory of distributions.
 
  • #5
Eh distribution theory is kind of overkill? I mean it's the most general way I've seen used to justify differentiating under the integral but in a lot of cases I've seen much more "low-tech" methods work as well.
 
  • #6
snipez90 said:
Eh distribution theory is kind of overkill? I mean it's the most general way I've seen used to justify differentiating under the integral but in a lot of cases I've seen much more "low-tech" methods work as well.
Well, you want to differentiate a distribution. You will be interchanging several limiting processes (the integral, the definition of the principal value and the differentiation itself) so you can't just differentiate under the integral sign and "expect" it to work.

However, using distributions the result is pretty much immediate.
 

What is the Cauchy residue formula?

The Cauchy residue formula is a powerful tool used in complex analysis to compute the value of a complex integral using the singularities of a function within a given contour. It states that the value of the integral is equal to the sum of the residues of the function at the singularities within the contour.

What is a residue in the context of the Cauchy residue formula?

In complex analysis, a residue is the coefficient of the term with a negative power of z in the Laurent series expansion of a function around a singularity. It is used to compute the value of a complex integral using the Cauchy residue formula.

What types of singularities can be used in the Cauchy residue formula?

The Cauchy residue formula can be applied to singularities of functions such as poles, essential singularities, and removable singularities. However, it cannot be applied to branch points or branch cuts.

Can the Cauchy residue formula be used to evaluate integrals on any contour?

Yes, the Cauchy residue formula can be used to evaluate integrals on any contour, as long as the contour encloses all the singularities of the function being integrated. If the contour encloses more than one singularity, the formula can be applied to each singularity separately and the results can be summed.

Are there any limitations to the Cauchy residue formula?

One limitation of the Cauchy residue formula is that it can only be applied to functions that are analytic within the contour of integration. Additionally, if the contour is not a simple closed curve, the formula may not give accurate results. It is also important to ensure that all singularities are included within the contour and that the contour does not intersect itself.

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