Simple integration question involving-infty subscript

  • Thread starter Thread starter Salt
  • Start date Start date
  • Tags Tags
    Integration
Click For Summary
SUMMARY

The discussion centers on the evaluation of the integral x(t) = ∫^t_{-\infty} x'(\tau) dτ, specifically addressing the misunderstanding of substituting infinity into a function. It is clarified that infinity is not a number and cannot be directly substituted, making the integral an improper integral. The correct approach involves using limits, expressed as x(t) = lim(a → -∞) ∫_a^t x'(\tau) dτ. This distinction is crucial for accurately solving problems involving improper integrals in calculus.

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with limits in calculus
  • Basic knowledge of signal processing concepts
  • Proficiency in calculus notation and operations
NEXT STEPS
  • Study the concept of improper integrals in calculus
  • Learn about limits and their application in integration
  • Explore signal processing fundamentals and their mathematical foundations
  • Practice solving integrals involving limits and improper integrals
USEFUL FOR

Students in calculus, particularly those studying signal processing, as well as educators and anyone looking to strengthen their understanding of improper integrals and limits in mathematical analysis.

Salt
Messages
15
Reaction score
0
simple integration question involving-infty "subscript"

Homework Statement



Been reading about signals, but my calculus skills have rusted (or never has been all that good in the first place).

So ...

Homework Equations



Why does [itex]x(t) = \int^t_{-\infty} x'(\tau) \,d\tau[/itex] ?

The Attempt at a Solution



You will end up with [itex]x(\tau)|^t_{-\infty} = x(t) - x({-\infty})[/itex]. Right?

So [itex]x({-\infty}) = 0[/itex] for all functions?

Been looking on the web, but I have no idea how to google this.
 
Physics news on Phys.org


Salt said:

Homework Statement



Been reading about signals, but my calculus skills have rusted (or never has been all that good in the first place).

So ...

Homework Equations



Why does [itex]x(t) = \int^t_{-\infty} x'(\tau) \,d\tau[/itex] ?

The Attempt at a Solution



You will end up with [itex]x(\tau)|^t_{-\infty} = x(t) - x({-\infty})[/itex]. Right?

So [itex]x({-\infty}) = 0[/itex] for all functions?

Been looking on the web, but I have no idea how to google this.
No, you can't do this. Infinity is not a number that you can substitute into a function. Your integral is one type of improper integral. To evaluate an integral like this, you need to work with a limit, like so:
[itex]x(t) = \int^t_{-\infty} x'(\tau) \,d\tau = \lim_{a \to -\infty} \int_a^t x'(\tau) \,d\tau[/itex]
 


Hmm ...

Looks like my calculus really does suck.

It was just written "like that" in the book. It probably assumes that I know how to solve it. ><

Thanks.
 
Last edited:

Similar threads

Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K