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Does this look right?

  1. Jul 11, 2006 #1
    [tex] \int_0^* \frac {dz}{z^23Z +2}[/tex]

    [tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{z^2+3Z+2}[/tex]

    [tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{(z+3/2)^2 -1/4}[/tex]

    let u=z+3/2

    [tex]=\lim_{t\rightarrow \*} \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

    [tex] = \frac{1}{2(1/2)}ln|\frac{u-1/2}{u+1/2}[/tex]

    [tex] = ln |\frac{z+1}{z+2}| \right]_0^*[/tex]

    [tex] = -ln \frac{1}{2}[/tex]

    this is one of the first of these im doing so bear with me if its horrible wrong :rolleyes:

    EDIT:* denotes infinite..PS is there a latex for infinite?
    Last edited: Jul 11, 2006
  2. jcsd
  3. Jul 11, 2006 #2
    I don't know why you used the lim symbol in the first place, and where the t commes in play, but

    [tex]= \int_0^* \frac{dz}{z^2+3Z+2} [/tex]
    [tex]= \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

    using your algebraic manipulations. After that, you used the partial sum decomposition skillfully and unless I'm mistaking, you are right!

    Just one minor thing : [tex] -ln \frac{1}{2} [/tex] = ln(2)
    Last edited: Jul 11, 2006
  4. Jul 11, 2006 #3
    yyyea..but what?I already had that is it right?
  5. Jul 11, 2006 #4


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    It's correct except you forgot to change the lower limit of integration. When z goes from 0 to infinity, then u = z + 3/2 goes from what to what?

    You do the infinity sign like [tex]\infty[/tex]
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