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Math Help - Integrate with respect to u

  1. #1
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    Integrate with respect to u

    Hi, I am having difficulty integrating the following with respect to u:

    e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c}

    Where c is a constant
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  2. #2
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    Separate the integrals, and then use integration by parts.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    EDIT: Wrong post... I thought that the solution could easily be obtained... my bad
    Last edited by Unknown008; July 4th 2010 at 04:56 AM.
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  4. #4
    Member mohammadfawaz's Avatar
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    Quote Originally Posted by Unknown008 View Post
    \int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} +c
    I think this is not correct. If we try to differentiate the right hand side, we don't get the function e^{f(x)}. For example, there is no closed form for \int e^{x^2}dx.
    To answer the question, note that  e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c} = e^c.e^{\frac{1}{2}u^2}.u^{-3/2}. Now, change your variable and continue (you may need to integrate by parts)
    But it gets very messy!
    http://www.wolframalpha.com/input/?i=integrate+e^(0.5*x^2-1.5*ln(x)%2Bc)dx++
    The answer is not encouraging :S
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    I think this is not correct. If we try to differentiate the right hand side, we don't get the function e^{f(x)}. For example, there is no closed form for \int e^{x^2}dx.
    To answer the question, note that  e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c} = e^c.e^{\frac{1}{2}u^2}.u^{-3/2}. Now, change your variable and continue (you may need to integrate by parts)
    But it gets very messy!
    http://www.wolframalpha.com/input/?i=integrate+e^(0.5*x^2-1.5*ln(x)%2Bc)dx++
    The answer is not encouraging :S
    Is that so? Sorry, I thought that it worked like the exponential functions I've been working with

    I tried on wolfram... it's the first time I see those symbols!

    http://www.wolframalpha.com/input/?i=\int+e^{\frac{1}{2}u^2+-+\frac{3}{2}\ln+u+%2B+c}+du
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  6. #6
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    Quote Originally Posted by Anne05 View Post
    Hi, I am having difficulty integrating the following with respect to u:

    e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c}

    Where c is a constant
    This is the same as Ce^{\frac{u^2}{2}u^{\frac{3}{2}} with C= e^c. If we let x=  \frac{u^2}{2}, then dx= u du and u^{\frac{3}{2}}du= (udu)(u^{\frac{1}{2}}= (2x)^{\frac{1}{4}}dx and \int e^{\frac{u^2}{2}}= 2^{\frac{1}{4}}\int x^{\frac{1}{4}}e^x dx
    I don't believe that can be integrated in terms of elementary functions.
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  7. #7
    Member mohammadfawaz's Avatar
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    Quote Originally Posted by Unknown008 View Post
    I tried on wolfram... it's the first time I see those symbols!

    http://www.wolframalpha.com/input/?i=\int+e^{\frac{1}{2}u^2+-+\frac{3}{2}\ln+u+%2B+c}+du
    Well the only mysterious symbol is \Gamma which is known as the Gamma function. It can be explained as a generalization of the factorial operator. It works on all real numbers. It is defined as integral.
    check it: Gamma Function -- from Wolfram MathWorld
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Yes, I was referring to that symbol, Gamma and another one, like 'O' in the series expansion of the integral. Thanks for the link! I don't think I'll understand these this year, maybe next year
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