nonelementary integrals

**Random Variable** · July 26th 2009, 06:35 PM

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?

**NonCommAlg** · August 15th 2009, 02:54 PM

Originally Posted by Random Variable

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?

**Random Variable** · August 15th 2009, 04:20 PM

Originally Posted by NonCommAlg

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 $\int e^{x^{2}} dx$ .

**NonCommAlg** · August 15th 2009, 04:35 PM

Originally Posted by Random Variable

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?

**mr fantastic** · August 15th 2009, 05:12 PM

Originally Posted by NonCommAlg

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).

**NonCommAlg** · August 15th 2009, 05:55 PM

thanks mr f.

Originally Posted by Random Variable

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.

**Random Variable** · August 16th 2009, 07:52 AM

I don't understand the case when q(x) is a polynomial of degree zero.

**Random Variable** · August 16th 2009, 11:36 AM

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

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

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

$f(x) = R'(x) +R(x)g'(x)$

**NonCommAlg** · August 16th 2009, 01:54 PM

Originally Posted by Random Variable

I don't understand the case when q(x) is a polynomial of degree zero.

$\deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.$

**Random Variable** · August 16th 2009, 04:16 PM

Originally Posted by NonCommAlg

$\deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.$

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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