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Simply as stated above: What is the proof, or how does one prove, $F(x)=\int_{a}^{b} e^{t^2}dt$ isn't elementary?

All I know is that it can be proven, but I couldn't find a proof for it.

VIVID
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Flechs
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    what do you mean by "elementary" ? – Surb Jul 28 '20 at 12:29
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    https://en.wikipedia.org/wiki/Risch_algorithm is an apropriate place to start. – 5xum Jul 28 '20 at 12:31
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    https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral and https://math.stackexchange.com/questions/287442/treatise-on-non-elementary-integrable-functions might be of interest – Matthew Towers Jul 28 '20 at 12:34
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    Look at Liouville theorem https://en.m.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) – EDX Jul 28 '20 at 12:43
  • See also https://math.stackexchange.com/a/2334/589 – lhf Jul 28 '20 at 12:44

2 Answers2

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Let's use the result of Liouville https://math.stackexchange.com/a/163/442

Let $f, g$ be rational functions, $g$ not constant. The indefinite integral $$ \int f(x)e^{g(x)}\;dx $$ is elementary if and only if there is a rational function $h$ so that $f=h'+hg'\;$.

Take $f(x) = 1$, $g(x) = x^2$ there. The theorem tells us that the indefinite integral $\int e^{x^2}\,dx$ is elementary if and only if there is a rational function $h(x)$ such that $f=h'+hg'\;$; that is: $$ 1 = h'(x)+2xh(x) . \tag1$$ Now if ($1$) holds in any interval of the real line, then it holds in all of $\mathbb C$.

In the linear differential equation with analytic coefficients ($1$), every point is a regular point. So the solution has no poles. The rational function $h$ is, in fact, a polynomial. Now let us study the behavior as $x \to \infty$. I claim $h(x)$ is bounded as $x \to \infty$. Suppose not. The Laurent series near $x=\infty$ is $$ h(x) = c x^m+O(x^{m-1})\qquad\text{as }x\to\infty $$ with $c\ne 0, m \ge 1$. So $$ h'(x) = cmx^{m-1}+O(x^{m-2}) $$ and $$ h'(x)+2xh(x) = 2 c x^{m+1} +O(x^{m}) $$ But $m \ge 1$ and $2c \ne 0$, so this is not the constant $1$.

Summary: $h$ is a polynomial, $h(x)$ is bounded as $x\to\infty$. Therefore $h$ is constant. Which again contradicts ($1$).

GEdgar
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More generally,

(A) Let $g$ be a polynomial. If $\int e^g$ is elementary, then $\deg{g}\le 1$.

This follows from a polynomial version of Liouville's theorem:

(B) Let $ f $ and $ g $ polynomials. If $ \int {f {e ^ g}} $ is elementary, then $ \int {f {e ^ g}} = P {e ^ g} $, where $ P $ is a polynomial.

This in turn follows from a rational version of Liouville's theorem:

(C) Let $ f $ and $ g $ be rational functions, with $ g $ not constant. If $ \int f {e ^ g} $ is elementary, then $ \int {f {e ^ g}} = R {e ^ g} $, where $ R $ is a rational function.

Theorem B implies theorem A:

By theorem B, with $ f = 1 $, we have $ \int e ^ g = P {e ^ g} $, where $ P $ is a polynomial. Taking derivatives and canceling $e^g$ gives $ 1 = P '+ Pg' $. If $ \deg {g}> 1 $, then $ \deg {g'} \geq 1 $ and so $ 0 = \deg {1} = \deg (P' + Pg') = \deg {Pg'} $, since $ \deg {P}> \deg {P'} $. Thus, $ Pg '$ is constant and so is $ g' $. But that implies $ \deg {g '} = 0 $ or $ g' = 0 $, a contradiction. So $ \deg {g} \le 1 $.

Theorem C implies theorem B:

By theorem C, we have $ \int {fe ^ g} = Re ^ g $, where $ R $ is a rational function. Taking derivatives and canceling $e^g$ gives $ f = R '+ Rg' $. Writing $ R = \frac {P} {Q} $ with $ P $ and $ Q $ relatively prime polynomials and $ Q $ monic, we have: $$ f = \frac {P'Q-PQ '} {Q ^ 2} + \frac {P} {Q} g' $$ and therefore $$ Q ^ 2f = P'Q-PQ '+ PQg' $$ and $$ Q (Qf-P'-Pg ') = -PQ' $$ Since $ P $ and $ Q $ are relatively prime, $ Q $ divides $ Q '$ and therefore $ Q = 1 $ (since $ Q $ is monic). So $ R = P $ and $ R $ is a polynomial.

lhf
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