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Exponential Integral question

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫(0 to 2) ∫(2x to 4) [e^(-y^2)] dydx

    This was a question given for a multivariable class exam a couple years ago, and we don't ever learn about the exponential integral in that class.

    2. Relevant equations

    We know how to go about solving this, BUT...

    3. The attempt at a solution
    So everyone knows you can't integrate e^(-y^2) dy directly. So i changed the bounds of integration to get ∫(0 to 4) ∫(y/2 to 2) [e^(-y^2)] dxdy. Doing that, the first integral became

    xe^(-y^2) from y/2 to 2, which gave me

    ∫(0 to 4) [2e^(-y^2) - ye^(-y^2)/2] dy

    So now I'm stuck with the same problem with that first part of the integral. Now what do I do?! Thanks.
     
  2. jcsd
  3. Dec 15, 2012 #2

    SammyS

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    Sketch the region of integration.

    You integrated over a different region. In the initial integration, y goes from 2x to 4, so in that region, y > 2x. If y > 2x, then x < y/2 .
     
  4. Dec 15, 2012 #3
    ok, done, fixed integral is now

    ∫(0 to 4) ∫(0 to y/2) [(-e^(-y^2))/4] dy. Same problem with the e^(-y^2) not being integratable.
     
  5. Dec 15, 2012 #4

    Ray Vickson

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    Your original integral was: first integrate y, then integrate x. You want to swap x and y, so you should first integrate x, then integrate y. If you sketch the integration region and take care you should get a simple final answer. If you don't take care you can certainly get into difficulties.
     
  6. Dec 15, 2012 #5

    lurflurf

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    You do not need a lesson for each integral. The e^(-y^2) is integrable. You might have meant the primitive function of e^(-y^2) is not an elementary function, which is unimportant. What is needed is to integrate by parts or interchange the integrals.

    by interchange the integrals
    [tex]\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=\int_0^4 \int_0^{y/2}e^{-y^2} \mathop{\text{dx dy}}[/tex]

    by integration by parts
    [tex]\int_0^2 \int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}=-\int_0^2 x \dfrac{d}{dx}\int_{2x}^4 e^{-y^2} \mathop{\text{dy dx}}[/tex]
     
  7. Dec 16, 2012 #6

    SammyS

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    @ lurflurf

    The anti-derivative of [itex]\displaystyle e^{-y^2}[/itex] is the error function. That's a special function. Therefore we often say that [itex]\displaystyle e^{-y^2}[/itex] is not integrable. Integration by parts does work for this function.


    @Hip2dagame:

    Integrating over x, [itex]\displaystyle \ \int_0^{y/2}e^{-y^2}\,dx\,,\ [/itex] treat the exponential, which is a function of y, as a constant. The result then allows integration by substitution, for the integration over y.
     
  8. Dec 16, 2012 #7

    lurflurf

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    That is a most unusual definition of integrable function, can you provide a reference to a book that uses that definition? The usual definition of integrable is that the integral should converge, you are referring to primitive functions that are not expressible in terms of elementary function. In this problem it is quite possible to avoid using the error function as I have above, it is also possible to use the error function when computing an elementary integral.

    [tex]\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2} [/tex]
     
  9. Dec 16, 2012 #8

    SammyS

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    You are correct regarding the term "integrable". I should have said the indefinite integral of ##\ \displaystyle e^{-y^2}\ ## cannot be expressed in closed form as a ordinary function -- and then mentioned the error function, erf(y).

    Regarding
    I appears that you are trying to use integration by parts to obtain the middle expression, the one involving the error function. That expression looks wrong to me. What steps are involved in obtaining that ?
     
  10. Dec 16, 2012 #9

    lurflurf

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    [tex]\int f(x)g ^\prime (x) \mathop{\text{dx}}=f(x)g(x)-\int f ^\prime (x)g(x)\mathop{\text{dx}}[/tex]
    where in this case
    f(x)=x
    g(x)=sqrt(pi)erf(x)
    and
    [tex]\text{erf}(x)=\dfrac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} \mathop{\text{dx}}[/tex]

    [tex]\int -2x e^{-x^2} \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2} [/tex]
    can be written
    [tex]\int -x (\sqrt{\pi} \text{erf}(x))^\prime \mathop{\text{dx}}=-x \sqrt{\pi} \text{erf}(x)+\int x^\prime \sqrt{\pi} \text{erf}(x) \mathop{\text{dx}}=\text{Constant}+e^{-x^2} [/tex]
     
  11. Dec 16, 2012 #10

    SammyS

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    It's certainly true that [itex]\displaystyle \int -2x e^{-x^2}\,dx=\text{Constant}+e^{-x^2}.\ [/itex] That result can arrived at through the substitution, u = -x2 .
     
  12. Dec 16, 2012 #11

    Ray Vickson

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    There is no need to integrate by parts.
    [tex] \int_{y=0}^4 \int_{x=0}^{y/2} e^{-y^2} \; dx \; dy
    = \int_{y=0}^4 \frac{y}{2} e^{-y^2} dy = -\frac{1}{4} \int_{y=0}^4 d(e^{-y^2}).[/tex]
     
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