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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

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    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

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    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.
     
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