Regularization by differentiation respect to a parameter

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SUMMARY

The discussion centers on the relationship between two integrals, I1 and I2, defined as I1 = ∫₀ⁿ dxx² (x-a)¹/² and I2 = ∫₀ⁿ dxx² (x-a)⁻¹/². It is established that I2 = 2(dI1/da) when applying a regularization scheme. The participants confirm that differentiating with respect to external parameters is generally valid, provided that the analytic dependence on the parameter 'a' is assumed. However, they emphasize the necessity for rigorous proofs that consider the specific properties of the chosen regularization scheme.

PREREQUISITES
  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with regularization techniques in mathematical analysis.
  • Knowledge of differentiation with respect to parameters in integrals.
  • Basic concepts of analytic functions and their properties.
NEXT STEPS
  • Study the properties of regularization schemes in mathematical physics.
  • Learn about analytic dependence and its implications in calculus.
  • Explore rigorous proofs related to differentiation under the integral sign.
  • Investigate advanced integral techniques, including contour integration.
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Mathematicians, physicists, and researchers involved in theoretical analysis, particularly those working with regularization methods and parameter differentiation in integrals.

zetafunction
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let be the integrals

[tex]\int_{0}^{\infty}dxx^{2} (x-a)^{1/2}=I1[/tex] and

[tex]\int_{0}^{\infty}dxx^{2} (x-a)^{-1/2}=I2[/tex]

is then correct that [tex]I2= 2\frac{dI1}{da}[/tex]

whenever applying a regularization scheme , is it correct to differentiate with respct to external parameters ??
 
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zetafunction said:
let be the integrals

[tex]\int_{0}^{\infty}dxx^{2} (x-a)^{1/2}=I1[/tex] and

[tex]\int_{0}^{\infty}dxx^{2} (x-a)^{-1/2}=I2[/tex]

is then correct that [tex]I2= 2\frac{dI1}{da}[/tex]

whenever applying a regularization scheme , is it correct to differentiate with respct to external parameters ??
Usually it is correct - typically one simply assumes an analytic dependence on a.

But rigorous proofs would have to take into account the detailed properties of a regularization scheme, and the meaning of the integrals...
 

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