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Homework Help: Integrate e^-a|x|

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm a little confused with this integral:


    2. Relevant equations
    3. The attempt at a solution

    Now, I believe the typical way to evaluate this is to say, hey, because of the |x|, this thing is symmetric about the x axis, and so we can instead evaluate:

    [tex]\int^\infty_{-\infty}e^{-a|x|}\,dx = 2\int^\infty_0e^{-ax}\,dx[/tex]
    [tex]= 2[-\dfrac{1}{a}e^{-ax}|^\infty_0] \, \, = \, 2[\dfrac{1}{a}] \, \, = \, \dfrac{2}{a}[/tex]

    which I believe is correct. However, and this is my question, can it be evaluated without using this trick? I ran into trouble, and I'm not sure where I made my mistake, although I suspect it has to do with not really doing anything about the absolute value of x:


    [tex] = \, \, [-\dfrac{1}{a}e^{-a|x|}|^\infty_{-\infty}][/tex]

    [tex] = \, \, -\dfrac{1}{a}[e^{-a|\infty|} - e^{-a|-\infty|}][/tex]

    [tex] = \, \, -\dfrac{1}{a}[e^{-a\infty} - e^{-a\infty}][/tex]

    [tex] = \, \, -\dfrac{1}{a}[0 - 0] \, = \, 0[/tex]

    And I've gotten nowhere, but I can't tell why, or what mistake I committed (if any).

    Thoughts? What am I missing here? Thanks!
  2. jcsd
  3. Oct 27, 2009 #2


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    Homework Helper

    the first way is correct it comes form the fact
    [tex]e^{-a|x|} = e^{-ax},x\geq 0[/tex]
    [tex]e^{-a|x|} = e^{ax},x < 0[/tex]

    so the integral becomes
    [tex]\int^\infty_{-\infty}e^{-a|x|}dx = \int^{0}_{-\infty}e^{ax}dx + \int^{\infty}_{0}e^{-ax}dx[/tex]

    which simplifies to what you gave (use substitution u = -x in first part)

    your 2nd interegral is not valid due to the different behaivour of [itex]e^{-a|x|} [/itex] either side of zero
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