How Can Divergent Integrals Be Handled with Ramanujan Summation?

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SUMMARY

Divergent integrals can be handled using Ramanujan summation, a technique that provides finite results for otherwise divergent series. The discussion specifically addresses the integrals of the forms \(\int_{0}^{\infty}dx \frac{x^{3}}{x+1}\) and \(\int_{0}^{\infty}dx \frac{x}{(x+1)^{1/2}}\). Participants explore the connection between Ramanujan summation and the summation of divergent series such as \(1 + 2 + 3 + 4 + 5 + 6 + 7 + \ldots\) and \(1 - 4 + 9 - 16 + 25\). The consensus is that Ramanujan summation offers a robust framework for assigning finite values to these divergent integrals.

PREREQUISITES
  • Understanding of divergent integrals
  • Familiarity with Ramanujan summation techniques
  • Knowledge of series convergence and divergence
  • Basic calculus, particularly integral calculus
NEXT STEPS
  • Research the principles of Ramanujan summation
  • Study methods for evaluating divergent integrals
  • Explore the implications of divergent series in mathematical analysis
  • Learn about other summation techniques such as Cesàro and Abel summation
USEFUL FOR

Mathematicians, physicists, and students interested in advanced calculus, particularly those dealing with divergent integrals and series. This discussion is beneficial for anyone looking to deepen their understanding of Ramanujan summation and its applications.

Klaus_Hoffmann
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Are there any method to deal with divergent integrals in the form

[tex]\int_{0}^{\infty}dx \frac{x^{3}}{x+1}[/tex] [tex]\int_{0}^{\infty}dx \frac{x}{(x+1)^{1/2}}[/tex] ?

in the same sense there are methods to give finite results to divergent series as 1+2+3+4+5+6+7+... or 1-4+9-16+25 or similar
 
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WHAT methods give finite results to 1+2+3+4+5+6+7+... and 1-4+9-16+25 ?
 
HallsofIvy said:
WHAT methods give finite results to 1+2+3+4+5+6+7+... and 1-4+9-16+25 ?

I've heard the term Ramanujan summation tossed about in regards to this, but I don't really know anything about it.
 

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