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Math Help - integration

  1. #1
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    integration

    What is the integal of e^(x^2)x^4 from negative infinity to infinity?

    Thank you!
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  2. #2
    Moo
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    Hello,

    The function is an even function ie f(x) = e^{x^2} x^4 = e^{(-x)^2} (-x)^4 = f(-x)

    ---------> \int_{- \infty}^{+\infty} f(x) dx = 2 \int_{-\infty}^0 f(x) dx

    Now, let's integrate.

    At first sight, you have to use the integration by parts, this must be almost a reflex when you see an exponential. The exponential will very often take the place of the v' function in :

    \int_a^b u(x)v'(x) dx = [u(x)v(x)]_a^b - \int_a^b u'(x)v(x) dx

    So you must make appear the derivate of x^2, 2x, so as to be able to integrate :

    the integrate of u'(x)e^{u(x)} will simply be e^{u(x)}

    2 \int_{-\infty}^0 e^{x^2} x^4 dx = \int_{-\infty}^0 (2x e^{x^2}) x^3 dx

    ------ integration by parts ------

    the derivate of x^3 is 3x^2

    \int_{- \infty}^{+\infty} e^{x^2} x^4 dx = [e^{x^2} x^3]_{-\infty}^0 - \int_{-\infty}^0 3x^2 e^{x^2} dx  = \lim_{x \to 0} e^{x^2} x^3 - \lim_{x \to - \infty} e^{x^2} x^3 - \int_{-\infty}^0 3x^2 e^{x^2} dx  = - \int_{-\infty}^0 3x^2 e^{x^2} dx

    (the two limits tend to 0)

    In the same way, make appear the derivate of x^2 so as to integrate once again by parts.

    - \int_{-\infty}^0 3x^2 e^{x^2} dx = - \frac{3}{2} \int_{-\infty}^0 (2x)e^{x^2} x dx = - \frac{3}{2} [e^{x^2} x]_{-\infty}^0 + \frac{3}{2} \int_{-\infty}^0 e^{x^2} dx


    And there, i can't help anymore because of the last integral : i don't remember how to do
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  3. #3
    Moo
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    Ay ! I found ^^


    This integrate is not defined, because though the limit in - \infty exists (0), the limit in  + \infty doesn't.

    \lim_{x \to +\infty} e^{x^2} x^4 = + \infty, so the integrate diverges.

    Are you sure it's the correct text ?
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  4. #4
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    Hi,

    Yea that is the question..ahhhhh I needed the answer to be zero! Ah well, thanks anyways!
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  5. #5
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    Note, x^2 \geq 0 \implies e^{x^2} \geq 1 \implies x^4 e^{x^2} \geq x^4.

    Thus, \int_{-\infty}^{\infty} x^4 e^{x^2} dx \geq \int_{-\infty}^{\infty}x^4 dx which clearly diverges since RHS diverges.

    Maybe the poster wanted, x^4 e^{-x^2}? (Note the negative).

    In that case,
    \int_{-\infty}^{\infty} x^4 e^{-x^2} dx = \int_{-\infty}^{\infty} \left( - \frac{1}{2} x^3 \right) \cdot \left( e^{-x^2} \right)' ~ dx =  -\frac{1}{2} x^3 e^{-x^2}\bigg|_{-\infty}^{\infty} +\int_{-\infty}^{\infty} \left(-\frac{3x}{4}\right)\cdot \left( e^{-x^2} \right) '  dx
    =-\frac{3}{4}xe^{-x^2} \bigg|_{-\infty}^{\infty} + \frac{3}{4}\int_{-\infty}^{\infty} e^{-x^2} dx = \frac{3\sqrt{\pi}}{8}
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  6. #6
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    Quote Originally Posted by Moo View Post
    Ay ! I found ^^


    This integrate is not defined, because though the limit in - \infty exists (0), the limit in  + \infty doesn't.

    \lim_{x \to +\infty} e^{x^2} x^4 = + \infty, so the integrate diverges.

    No, \lim_{x \to - \infty}e^{x^2} x^4 = +\infty too.
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