Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definite integrals with -infinity low bound

  1. Feb 21, 2012 #1
    I see equations of the form,

    [itex]y=\int_{-\infty }^{t}{F\left( x \right)}dx[/itex]

    a lot in my texts.

    What exactly does it mean? From the looks of it, it just means there is effectively no lower bounds.

    I looked up improper integrals, but I can't say I really understand what is going on.

    So when evaluating,

    If [itex]\frac{d\left( f\left( x \right) \right)}{dx}=F\left( X \right)[/itex]

    Do I just take the lower bound term - that I have to subtract - to be the f(x) as x approaches -infinity? Do I set the lower bound term to 0?

    [itex]y=\int_{-\infty }^{t}{F\left( x \right)}dx\; =\; f\left( t \right)-0=f\left( t \right)[/itex]

    ?
     
  2. jcsd
  3. Feb 21, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's defined this way:

    [tex]\int_{-\infty}^t F(x) dx = \lim_{a \rightarrow -\infty} \int_{a}^{t} F(x) dx[/tex]

    and you would evaluate it as follows:

    [tex]y = \int_{-\infty}^t F(x) dx = \lim_{a \rightarrow -\infty} \int_{a}^{t} F(x) dx = f(t) - \lim_{a \rightarrow -\infty} f(a)[/tex]
     
  4. Feb 21, 2012 #3
    OK I just saw this in my text (paraphrasing),

    [itex]y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau [/itex]

    therefore

    [itex]x\left( t \right)=\frac{dy}{dt}[/itex]​

    So [itex]y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau [/itex] is just like a non-definite integral?

    [itex]y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau =\; \int_{}^{}{x\left( t \right)dt}[/itex] ?

    BTW I'm studying an engineering text, so maybe it's "shortcuted" somehow.
     
  5. Feb 21, 2012 #4

    Char. Limit

    User Avatar
    Gold Member

    I think it's pretty easily provable that in order for a definite integral with a lower bound at -infinity to exist, the function being integrated needs to tend to zero as the variable being integrated tends to -infinity. (Note that it doesn't work the other way! Try -1/x for proof!) So the derivative of the integral would be just the integrand, assuming that the integrand satisfies the conditions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Definite integrals with -infinity low bound
  1. Definite integral (Replies: 7)

Loading...