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I'm trying this exercise:

Show that, for each rational function $h(x)$, the function $f(x)=e^{x^2}$ doesn't admit (on any interval in $\mathbb{R}$) a primitive in the form $G(x)=e^{x^2}h(x)$.

I suppose that $G'(x)=f(x) \ \ \ \rm \forall x \in I \subset \mathbb{R}$ for a interval, so $G'(x)=e^{x^2}h'(x) + h(x)e^{x^2}2x$. But $h(x)=\dfrac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials, so $G'(x)=e^{x^2}\dfrac{p'(x)q(x)-q'(x)p(x)}{(q(x))^2}+\dfrac{2xe^{x^2}p(x)}{q(x)}=e^{x^2}$. Simplifying the equation I get $p'(x)q(x)-q'(x)p(x)+2xp(x)q(x)=(q(x))^2$. Now, I would like to find a contradiction, but I can't. Perhaps the I interpreted wrongly the text and the polynomials $p(x)$ and $q(x)$ are integer polynomials?

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  • $\begingroup$ Should $f(x)$ have been $e^{-x^2}$ instead? $\endgroup$ Sep 7, 2017 at 19:08
  • $\begingroup$ I think you are on the right track, and $p,q$ might have real coefficients (or complex). The key is that $p,q$ will have no common factor, and.derive a contradiction from that. $\endgroup$
    – hardmath
    Sep 7, 2017 at 19:15
  • $\begingroup$ @henning-makholm No, $f(x)=exp(x^2)$. $\endgroup$ Sep 7, 2017 at 19:15
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    $\begingroup$ If $\deg p\ge\deg q$, then you should be able to show that $2xp(x)q(x)$ has a higher degree than any other term in that identity you derived. This is a contradiction, do you see why? $\endgroup$ Sep 7, 2017 at 19:16
  • $\begingroup$ A similar contradiction is available if $\deg q>\deg p+1$. Then $q(x)^2$ has higher degree than any of the others. This leaves as the only possibility that $\deg q=\deg p+1$, and the highest degree terms of $2xp(x)q(x)$ and $q(x)^2$ cancel each other from the identity. $\endgroup$ Sep 7, 2017 at 19:20

1 Answer 1

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A way to wrap this up is the following. Alas, this assumes that you know a few things about divisibility of polynomials.

Without loss of generality we can assume that $q(x)$ and $p(x)$ have no common factors. For we could always simply cancel that factor.

I first claim that $q(x)$ must be constant. Assume not. Then it has some irreducible polynomial $r(x)$ as a factor (so either $r(x)$ is linear, or a quadratic without real zeros). Let $n$ be the highest power such that $r(x)^n\mid q(x)$. So $n\ge1$. Then

  • $r(x)^{2n}$ is a factor of the right hand side.
  • the polynomials $2xp(x)q(x)$ and $q(x)p'(x)$ are both divisible by $r(x)^n$.
  • because the polynomial $p(x)$ is not divisible by $r(x)$, the product $p(x)q'(x)$ is divisible by $r(x)^{n-1}$ but not by $r(x)^n$.

Therefore in your equation there is only one term that is not divisible by $r(x)^n$. This is a contradiction.

Ok. So $q(x)$ is a constant. Without loss of generality $q(x)=1$. Then the polynomial $2xp(x)$ has higher degree than any other term. That is also a contradiction. Done.

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  • $\begingroup$ This is somehow unsatisfactory. I tend to think that something simpler is out there. Like, can we prove that $h(x)$ must "asymptotically" be increasing? That would imply $\deg p\ge\deg q$, and we are done by degree comparisons alone. OTOH asymptotic considerations won't apply on some finite interval (unless we do something like analytic continuation). $\endgroup$ Sep 7, 2017 at 21:35
  • $\begingroup$ It seems to me that something is wrong somewhere..r(x) could divide q'(x) for example take the derivative of $(x-3)^3 $ well it's derivative could still be divide by (x-3) ... if r'x) divide q(x) we can say nothing about q'(x)...does r(x) divide q' or not ? Thats not impossible...Am I wrong ? $\endgroup$ Sep 8, 2017 at 0:50
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    $\begingroup$ @Isham: Here $q(x)=r(x)^ns(x)$ for some polynomial $s(x)$ not divisible by $r(x)$. Therefore $q'(x)=nr(x)^{n-1}s(x)r'(x)+r(x)^ns'(x)$. Here the first term is divisible by $r(x)^{n-1}$ but not by $r(x)^n$. The latter term is divisible by $r(x)^n$. Putting these together we see that $q'(x)$ is divisible by $r(x)^{n-1}$ but not by $r(x)^n$. A subtlety is that $r'(x)s(x)$ cannot be divisible by $r(x)$. Neither $r'(x)$ nor $s(x)$ is, and because $r(x)$ is irreducible the product cannot be either. $\endgroup$ Sep 8, 2017 at 4:48
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    $\begingroup$ In other words $q'(x)$ is divisible by $r(x)^{n-1}$. The point is that it is not divisible by $r(x)^n$. Consequently neither is $p(x)q'(x)$. $\endgroup$ Sep 8, 2017 at 4:52

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