Is the Laplace Transform F(s) = 1/(1+s^2) Defined for All Real Values of s?

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SUMMARY

The Laplace Transform F(s) = 1/(1+s^2) is indeed defined for all real values of s. This conclusion arises from the fact that the expression 1/(1+s^2) does not encounter any singularities for any real number input, as s^2 is always non-negative. Therefore, F(s) remains valid regardless of whether s is greater than or less than zero.

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  • Understanding of Laplace Transforms
  • Basic knowledge of complex analysis
  • Familiarity with real-valued functions
  • Concept of singularities in mathematical functions
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  • Study the properties of Laplace Transforms in detail
  • Explore the implications of singularities in complex functions
  • Learn about the inverse Laplace Transform techniques
  • Investigate applications of Laplace Transforms in differential equations
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Students of engineering and mathematics, particularly those studying control systems or differential equations, will benefit from this discussion on the validity of the Laplace Transform across all real values of s.

ming2194
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Homework Statement


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Homework Equations


find above

The Attempt at a Solution



in fact i had no problem in the progress of doing this question and finally i got F(s)= 1/1+s^2

and i want to ask it is vaild for all value of s since it is s^2?
is it true for no matter s is >0 or s is <0?
 
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I can't think of any reason why it wouldn't be defined for all real s.
 

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