Simple integration question involving-infty subscript

  • Thread starter Thread starter Salt
  • Start date Start date
  • Tags Tags
    Integration
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
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: