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We can't find $$ \int e^{t^2} \; dt $$ using basic tools from a calculus class. That is, we can't express an antiderivative of $f(t) = e^{t^2}$ using the basic operations. We can of course just define $$ F(t) = \int_{a}^t e^{s^2}\; ds. $$

I am looking for a way to explain to a student why we can't express the antiderivative using basic operations (addition, subtraction, root, powers, etc.). In particular I am interested in getting at answering the "why we can't do that". I know that one could "just" prove it, but is there a simple argument or something that illustrates this?

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2 Answers 2

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First, it is not that we cannot find an antiderivative: we can, in fact: the function $$F(t)=\int_0^t\exp(s^2)\,\mathrm d s$$ is a perfectly good antiderivative. What is true is that you cannot find an expression of an antiderivative as an elementary function (for some specific meaning of «elementary function»)

Now, to prove this one needs to make precise what is meant by elementary and then prove it. The proof is not something you'd be able to present to a student learning integration, sadly. I guess there are several ways to do it, but the standard argument involves the theory of differential field extensions, as explained in the little book by Andy Majid on the subject.

The argument is not unsimilar to that which shows that there is no general formula for the roots of a polynomial —this gives you something to compare it with.

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    $\begingroup$ Yes, I know that an antiderivative is defined like you write, but I wanted to get at the "why can't we find this like we can find other antiderivatives using tools like substitutions" $\endgroup$
    – Thomas
    Apr 10, 2013 at 21:54
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    $\begingroup$ +1: I presume this is the book: "Magid, Andy R., Lectures on differential Galois theory". $\endgroup$
    – copper.hat
    Apr 10, 2013 at 22:00
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    $\begingroup$ Exactly! ${}{}$ $\endgroup$ Apr 10, 2013 at 22:00
  • $\begingroup$ The "general formula for the roots of a polynomial"-problem was solved beautifully by Galois theory; is this problem equally elegantly solved? $\endgroup$
    – akkkk
    Apr 10, 2013 at 22:16
  • $\begingroup$ You may also google Liouville's theorem. French mathematician Liouville descended upon this and it gives a condition when such an e-power is integrable (for example, if it has a linear term in front, but there is more to that). While it may not be "classroom material", it certainly helped me to understand. $\endgroup$
    – imranfat
    Apr 10, 2013 at 22:28
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You can't prove it easily, but you can give some intuition :

  1. derivative of $e^x$ is $e^x$
  2. Trying to find $f$ such that $f'(x)=e^{x^2}$, we can guess that $f(x)=g(x)e^{h(x)}$
  3. So $f'(x)=(g'(x)+g(x)h'(x))e^{h(x)}=e^{x^2}$
  4. So we want to find a solution such that $h(x)=x^2$ and $g'(x)+2xg(x)=1$

Note the last point seems easier, but if they try, they will not find elementary function $g$ such that $g(x)=\frac{1-g'(x)}{2x}$. You can at least show this is not a polynomial function, nor a quotient of polynomials function, nor a usual trigonometric function.

But let $g(x)=\sum a_ix^i$, then $a_1=1$ and $(i+1).a_{i+1}+2a_{i-1}=0$, so you can at least give some expression to $g$ by solving $a_i$ (you can suppose that $a_0=0$).

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