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Math Help - Integral

  1. #1
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    Integral

    Compute \int_{-1}^1\frac{x^2}{1+e^x}\,dx

    I can't get it!
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  2. #2
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    Facts you should know

    \frac1{1+e^x}+\frac1{1+e^{-x}}=1

    Set u=-x
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Facts you should know

    \frac1{1+e^x}+\frac1{1+e^{-x}}=1

    Set u=-x
    Clever. No Physicist ever taught me that kind of trick when doing this integral. (It crops up in integrals over fermionic particles in solid state Physics.)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Facts you should know

    \frac1{1+e^x}+\frac1{1+e^{-x}}=1

    Set u=-x
    ...umm...i don't get it. how do we apply his exactly?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by liyi View Post
    Compute \int_{-1}^1\frac{x^2}{1+e^x}\,dx

    I can't get it!
    Quote Originally Posted by Jhevon View Post
    ...umm...i don't get it. how do we apply his exactly?
    Look at the u = -x case:
    \int_1^{-1}\frac{(-u)^2}{1+e^{-u}}\,(-du)

    = \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx

    Now add the two integrals. (Hint: Do an integration by parts on the second integral and see what you get.)

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Facts you should know

    \frac1{1+e^x}+\frac1{1+e^{-x}}=1

    Set u=-x
    On second thought this trick isn't likely to work well in Physics. The reason this works here is because of the symmetric limits, and the Physics integrals go from 0 to \infty.

    It's still a really cool way to do this integral, though.

    -Dan
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Look at the u = -x case:
    \int_1^{-1}\frac{(-u)^2}{1+e^{-u}}\,(-du)

    = \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx

    Now add the two integrals. (Hint: Do an integration by parts on the second integral and see what you get.)

    -Dan
    ok, i'm not getting anywhere with this:

    \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx = -x^2 \ln \left( e^{-x} + 1 \right) + 2 \int x \ln \left( e^{-x} + 1 \right) ~dx

    i also did by parts on the last integral also, but it never seemed to be leading me anywhere. and even if i could get a nice result, i'm not exactly sure how you expect me to combine the two integrals when they have different variables.
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx
    This is the key, 'cause we have the same integration limits, so we can turn back the u into x

    Now

    \int_{-1}^1\frac{x^2}{1+e^x}\,dx+\int_{-1}^1\frac{x^2}{1+e^{-x}}\,dx=\frac13x^3\bigg|_{-1}^{\phantom{-}1}=\frac23

    Divide by two this last result and yields the desired.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, i'm not getting anywhere with this:

    \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx = -x^2 \ln \left( e^{-x} + 1 \right) + 2 \int x \ln \left( e^{-x} + 1 \right) ~dx

    i also did by parts on the last integral also, but it never seemed to be leading me anywhere. and even if i could get a nice result, i'm not exactly sure how you expect me to combine the two integrals when they have different variables.
    Hmmm...

    I plugged the integral in on my calculator and got that
    \int \frac{x^2}{1 + e^{-x}}dx = \frac{1}{3}x^3 - \int \frac{x^2}{1 + e^x} dx

    At a glance it looked like an integration by parts, but I guess it isn't. (Or, rather, it still may be but I can't figure how to do it.) I'll have to work on the integral a bit, but in any event you should be able to see where to go with it from here.

    Edit: Ah wait! I see where Krizalid is going with this. The integrand isn't an even function, but the two are reflections of each other over the y-axis. Since the integration limits are symmetric the two integrals must be equal to each other.

    -Dan
    Last edited by topsquark; September 21st 2007 at 05:49 PM. Reason: Second thoughts
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    This is the key, 'cause we have the same integration limits, so we can turn back the u into x

    Now

    \int_{-1}^1\frac{x^2}{1+e^x}\,dx+\int_{-1}^1\frac{x^2}{1+e^{-x}}\,dx=\frac13x^3\bigg|_{-1}^{\phantom{-}1}=\frac23

    Divide by two this last result and yields the desired.
    Yes! that was the key! Got it! now i really appreciate the ingenious of your substitution, i wouldn't thought of that in a million years! how did you "see" that?!
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  11. #11
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    It is just practice.

    Since I like integrals too
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    It is just practice.
    no, it's because you are the man!!!!
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