Does the Integral \(\int_0^\infty \sin(x) \, dx\) Have a Definite Value?

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Discussion Overview

The discussion centers around the improper integral \(\int_0^\infty \sin(x) \, dx\) and its potential to have a definite value. Participants explore the conditions under which such integrals exist, particularly in the context of a related integral involving an exponential function.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions whether the integral \(\int_0^\infty \sin(x) \, dx\) has a definite value.
  • Another participant asks about the requirements for an improper integral to exist, indicating uncertainty about the well-defined nature of the integral.
  • A participant clarifies that the original integral in question is actually \(\int_0^\infty \sin(x) e^x \, dx\), suggesting that this form may be more problematic.
  • One participant proposes that a missing minus sign in the exponent could lead to the existence of the integral, implying that the sign is crucial for convergence.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the existence of the integral, with some suggesting conditions that could lead to a well-defined value while others remain skeptical. No consensus is reached on the matter.

Contextual Notes

The discussion highlights the importance of the exponent's sign in determining the convergence of the integral, but the specific conditions and implications remain unresolved.

jollage
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Hi

Does this integral have a definite value [itex]\int^{\infty}_0 sin(x)dx[/itex]?

Jo
 
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What is the requirement for an improper integral of this type to exist?
 
arildno said:
What is the requirement for an improper integral of this type to exist?

Thanks. So you imply it's not a well-defined integral. Actually the original integral is [itex]\int^{\infty}_0 sin(x) e^x dx[/itex]. This original integral comes from a step during my derivation of the solution to the IBVP of Burger's equation. This original one seems more absurd, isn't it?

Jo
 
Sure.
Most likely, you have missed a minus sign in the exponent, in which case the integral will exist.
 
arildno said:
Sure.
Most likely, you have missed a minus sign in the exponent, in which case the integral will exist.

I see your point. Thanks.

Jo
 

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