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'trick' of differentiating

  1. Oct 11, 2006 #1
    [tex] \int_{-\infty}^{infty} s e^{-\frac{2s^2}{N}} ds [/tex]

    how do i integrate here?? I dont think the 'trick' of differentiating wrt N would work here since the limits of integration are all space....

    any ideas??
     
  2. jcsd
  3. Oct 11, 2006 #2

    StatusX

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    substitution
     
  4. Oct 19, 2006 #3
    ok this is substitution i did

    let x^2 = u
    then 2xdx = du

    [tex] I = \frac{1}{2\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{\frac{-u}{2 \sigma^2}} du [/tex]

    what happens in [tex] \left[ e^{-u} \right]_{-\infty}^{\infty} \rightarrow \infty [/tex]

    something isnt right ...?

    its supposed to be zero, no?
     
  5. Oct 20, 2006 #4

    dextercioby

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    The Cauchy principal value of that integral is zero and that can be seen since you're integrating an odd function on an interrval symmetric wrt zero on the real axis.

    Daniel.
     
  6. Oct 20, 2006 #5

    HallsofIvy

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    When you change variables, change the limits of integration too.
    Setting u= x2 in
    [tex] \int_{-\infty}^{\infty} s e^{-\frac{2s^2}{N}} ds [/tex]
    (I would have been inclined to let u be the whole [itex]\frac{2s^2}{N}[/itex].)
    does NOT give
    [tex] I = \frac{1}{2\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{\frac{-u}{2 \sigma^2}} du [/tex]
    you have the wrong limits of integration.

    Actually, you don't need to use substitution at all. The integral of any odd function from -A to A is what?
     
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