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Wierd Integral

  1. Apr 8, 2008 #1
    Why does this work? MainEq1.L.gif

    Or maybe a better question is how do you evaluate this integral? Integrate a piecewise function? I tried that an got 0
  2. jcsd
  3. Apr 8, 2008 #2


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    You subtracted when you should've added. It's definitely 2.
  4. Apr 8, 2008 #3
    I get [tex]-e^{-x}, x> 0 -- e^x, x<0[/tex] But I'm unsure as how to go about solving that.

    I evaluated it like this:

    [tex]-e^{-x}|_{x=inf} - e^{x}|_{x=-inf}[/tex] but that's clearly wrong XD
    Last edited: Apr 8, 2008
  5. Apr 8, 2008 #4
    an Improper integral

    you have to Divide the integral in to two
    the first is from -∞ to 0 & the second is from 0 till ∞ and
    HINT: this is an improper integral, consider taking the limit
    for example limC goes to -∞ (of your integral)
    lim D goes to ∞ (of your integral).
    and continue.:smile:
  6. Apr 8, 2008 #5
    [tex]\int_{-\infty}^{\infty}e^{-|t|}dt=\lim_{a\rightarrow -\infty}\int_{a}^{c}e^{-|t|}dt+\lim_{b\rightarrow \infty}\int_{c}^{b}e^{-|t|}dt[/tex]


    [tex] e^{-|t|}=e^{-t}, t>0[/tex] and [tex] e^{t},t<0[/tex]
    YOu can choose c to be any point between negative infinity and positive infinity. Let c=0 so

    [tex]\int_{-\infty}^{\infty}e^{-|t|}dt=\lim_{a\rightarrow -\infty}\int_{a}^{0}e^{t}dt+\lim_{b\rightarrow \infty}\int_{0}^{b}e^{-t}dt[/tex]
  7. Apr 8, 2008 #6
    Yeah I figured it out earlier today. You can do that method or since the function is even throw out the absolute value and evaluate the integral from 0 to infinity and multiply by 2.
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