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Tough integration

  1. Sep 27, 2008 #1
    integral of x*(1/2b)*exp(-abs(x-a)/b)

    sorry about the format, I dont know how to use the signs.

    this looks like an integration by parts, but I'm not really seeing how to work it out

    thanks!
     
  2. jcsd
  3. Sep 27, 2008 #2

    arildno

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    Would that be something like:
    [tex]\int{x}\frac{1}{2b}e^{-\frac{|x-a|}{b}}dx[/tex]??
     
  4. Sep 27, 2008 #3
    Exactly. Thanks.
     
  5. Sep 27, 2008 #4

    arildno

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    Can you pinpoint the problem or problems you have in finding an anti-derivative to that function?
     
  6. Sep 27, 2008 #5

    arildno

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    A helpful advice is to look at the DEFINITION of the absolute value function:
    |y|=y if y>=0 and |y|=-y if y<0

    Thus, for x-a>=0 (i.e, x>=a), we may simplify:
    [tex]\int{x}\frac{1}{2b}e^{-\frac{|x-a|}{b}}dx=\int{x}\frac{1}{2b}e^{-\frac{x-a}{b}}dx=\frac{1}{2}e^{\frac{a}{b}}\int\frac{x}{b}e^{-\frac{x}{b}}dx=\frac{b}{2}\int{u}e^{-u}du,u=\frac{x}{b}[/tex]
    This is fairly trivial to anti-differentiate.

    Make similar simplifications for the case x-a<0.
     
  7. Sep 27, 2008 #6
    I'm not really sure how to deal with the absolute value
     
  8. Sep 27, 2008 #7
    oh ok thanks
     
  9. Sep 27, 2008 #8
    how would I got about finding the value of the integral from negative infinity to positive infinity if I have 2 different expressions depending on the value of x-a?
     
  10. Sep 27, 2008 #9
    nevermind, i got it
     
  11. Sep 27, 2008 #10
    I haven't dealt with improper integrals yet, but my guess would be to integrate it over two separate intervals. One from negative infinite to 0 and the other from 0 to positive infinite.
     
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