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Wierd integration of exp(-t²)

  1. May 18, 2012 #1
    Dear All,

    I computed an integral that looks like erf(x) without problem: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \int_{-\infty}^{0} e^{-t^2} dt + \int_{0}^{+\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} [-erf(-\infty)+erf(+\infty)] = \frac{\sqrt{\pi}}{2} [-(-1)+1] = \sqrt{\pi}. [/tex]

    However, what about the change of variable: [tex] u = t^2? [/tex]
    Hence: [tex] du = 2 dt [/tex] and: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \frac{1}{2} \int_{+\infty}^{+\infty} e^{-u} du = -\frac{1}{2} [e^{-u}]_{+\infty}^{+\infty} = 0. [/tex]

    I just want to be sure: the second integration doesn't make sense, because the lower and upper bounds are the same, right?

    Thanks.

    Regards.
     
  2. jcsd
  3. May 18, 2012 #2
    This is exactly why one splits up an integral from -∞ to ∞ into two parts and takes the limit of each: the normal rules of integration don't necessarily apply.
     
  4. May 18, 2012 #3
    No

    du=2tdt
     
  5. May 18, 2012 #4
    Indeed, my bad, thanks to both of you.
     
  6. May 18, 2012 #5

    D H

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    You've already been told about the problem with du. Another problem here: u cannot be negative, yet you have the integrations limits being -∞ to +∞.
     
  7. May 18, 2012 #6
    Dear D H,

    Thanks for your reply. It's now clear to me that this change of variable is for positive values only.

    Regards.
     
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