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Math Help - nonelementary integrals

  1. #1
    Super Member Random Variable's Avatar
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    nonelementary integrals

    There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then \int f(x)e^{g(x)} \ dx can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that  f(x)=R'(x)+R(x)g(x) .


    So let's take \int \frac{\sin x}{x} \ dx = \int \sin x \ e^{- \ln x} \ dx

    So I have to show that there does not exist a R(X) such that  \sin(x)=R'(x)-R(x)\ln x .

    It's a linear first order differential equation with integrating factor e^{-x \ln x+x} = x^{-x}e^{x}

     x^{-x}e^{x}R'(x) - x^{-x}e^{x}\ln x R(x) = x^{-x}e^{x}\sin x

     \frac{d}{dx} \int \big(x^{-x}e^{x}R(x)\big) = \int x^{-x}e^{x}\sin x \ dx

     R(x) = x^{x}e^{-x} \int^{x}_{0} t^{-t}e^{t}\sin t \ dt + Cx^{x}e^{-x}

    How do I show that R(x) is not rational? Do I have to show that  \int x^{-x}e^{x}\sin x \ dx is nonelementary?
    Last edited by Random Variable; August 14th 2009 at 03:59 PM.
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    Quote Originally Posted by Random Variable View Post
    There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then \int f(x)e^{g(x)} \ dx can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that  f(x)=R'(x)+R(x)g(x) .


    So let's take \int \frac{\sin x}{x} \ dx = \int \sin x \ e^{- \ln x} \ dx

    So I have to show that there does not exist a R(X) such that  \sin(x)=R'(x)-R(x)\ln x .

    It's a linear first order differential equation with integrating factor e^{-x \ln x+x} = x^{-x}e^{x}

     x^{-x}e^{x}R'(x) - x^{-x}e^{x}\ln x R(x) = x^{-x}e^{x}\sin x

     \frac{d}{dx} \int \big(x^{-x}e^{x}R(x)\big) = \int x^{-x}e^{x}\sin x \ dx

     R(x) = x^{x}e^{-x} \int^{x}_{0} t^{-t}e^{t}\sin t \ dt + Cx^{x}e^{-x}

    How do I show that R(x) is not rational? Do I have to show that  \int x^{-x}e^{x}\sin x \ dx is nonelementary?
    this is a good question! i just saw it but, considering your last post, i'm not sure if you're still looking for a solution or not?
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this is a good question! i just saw it but, considering your last post, i'm not sure if you're still looking for a solution or not?
    Yes, I would still like a solution. I just thought that the example I gave doesn't apply because  \ln(x) isn't a rational function.

    But I run into the same issue with  \int e^{x^{2}} dx .
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then \int f(x)e^{g(x)} \ dx can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that  f(x)=R'(x)+R(x)g(x). \ \ \ \color{red}(*)
    i think \color{red} (*) is not correct. shouldn't it be f(x)=R'(x)+R(x)g'(x) instead?
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    i think \color{red} (*) is not correct. shouldn't it be f(x)=R'(x)+R(x)g'(x) instead?
    Correct. There is a typo at the source the OP referenced the result from (the typo is very old now and was never corrected by the webmaster).
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  6. #6
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    thanks mr f.

    Quote Originally Posted by Random Variable View Post
    But I run into the same issue with  \int e^{x^{2}} dx .
    let's prove a more general claim: if g(x) is a polynomial and \deg g(x) \geq 2, then \int e^{g(x)} \ dx cannot be expressed in terms of a finite number of elementary functions.

    suppose otherwise. then there exists a rational function R(x)=\frac{p(x)}{q(x)}, where p(x),q(x) are some polynomials in \mathbb{C}[x] and R'(x)+g'(x)R(x) = 1. obviously we may assume that p(x),q(x)

    have no common zeros. we have R'(x)=\frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}. thus: p'(x)q(x)-p(x)q'(x)+g'(x)p(x)q(x)=q(x)^2, which gives us: q(x)[p'(x)-q(x)+g'(x)p(x)]=p(x)q'(x). \ \ \ \ \ (1)

    now consider two cases:

    case 1: \deg q(x)=0: in this case R(x) would be a polynomial satisfying R'(x)+g'(x)R(x)=1. this is impossible because \deg(R'(x)+g'(x)R(x)) \geq 1 + \deg R(x) \geq 1.

    case 2: \deg q(x) \geq 1. let q(x)=(x-a)^kq_1(x), where k \geq 1 and q_1(a) \neq 0. put this in (1) to get: (x-a)q_1(x)[p'(x)-q(x)+g'(x)p(x)]=kp(x)q_1(x)+(x-a)p(x)q_1'(x), which is impossible

    because putting x=a will give us p(a)=0 and so x=a would be a common zero of p(x),q(x). contradiction! Q.E.D.
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  7. #7
    Super Member Random Variable's Avatar
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    I don't understand the case when q(x) is a polynomial of degree zero.
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  8. #8
    Super Member Random Variable's Avatar
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    Is it equivalent to say that if f(x) and g(x) are rational functions,  \int f(x)e^{g(x)} can be expressed in terms of a finite number of elementary functions iff it's antiderivative is of the form  R(x)e^{g(x)} where R(x) is a rational function?


    Because then  \int f(x)e^{g(x)} = R(x)e^{g(x)}

     \frac{d}{dx} \int f(x)e^{g(x)} = \frac{d}{dx} R(x)e^{g(x)}

     f(x)e^{g(x)} = R'(x)e^{g(x)} + R(x)g'(x)e^{g(x)}

     f(x) = R'(x) +R(x)g'(x)
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  9. #9
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    Quote Originally Posted by Random Variable View Post
    I don't understand the case when q(x) is a polynomial of degree zero.
    \deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.
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  10. #10
    Super Member Random Variable's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    \deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.
    Thanks.


    Is it possible to write  \frac {\sin x}{x} and/or  x^{x} so that they are in the proper form to use the theorem?

    My intuition tells me to use the complex form of  \sin x for  \frac{\sin x}{x} . But then I have the difference of two functions of the proper form.

    I have no idea what to do with  x^{x} . Unless my understaning of the definition of a rational function is incorrect (namely that a rational function is function that can be written as the quotient of two polynomials) I can't writex  x^{x} as e^{x \ln x}
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