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We know that the derivative of some non-elementary functions can be expressed in elementary functions. For example $ \frac{d}{dx} Si(x)= \frac{\sin(x)}{x} $

So similarly are there any non-elementary functions whose integrals can be expressed in elementary functions?

If not then how can we prove that any integral in the set of non-elementary integrals cannot be expressed in the form of elementary functions?

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The derivative of an elementary function is an elementary function: the standard Calculus 1 differentiation methods can be used to find this derivative. So an antiderivative of a non-elementary function can't be elementary.

EDIT: More formally, by definition an elementary function is obtained from complex constants and the variable $x$ by a finite number of steps of the following forms:

  1. If $f_1$ and $f_2$ are elementary functions, then $f_1 + f_2$, $f_1 f_2$ and (if $f_2 \ne 0$) $f_1/f_2$ are elementary.
  2. If $P$ is a non-constant polynomial whose coefficients are elementary functions, then a function $f$ such that $P(f) = 0$ is an elementary function.
  3. If $g$ is an elementary function, then a function $f$ such that $f' = g' f$ or $f' = g'/g$ is elementary (this is how $e^g$ and $\log g$ are elementary).

To prove that the derivative of an elementary function, you can use induction on the number of these steps. In the induction step, suppose the result is true for elementary functions obtained in at most $n$ steps. If $f$ can be obtained in $n+1$ steps, the last being $f = f_1 + f_2$ where $f_1$ and $f_2$ each require at most $n$ steps, then $f' = f_1' + f_2'$ where $f_1'$ and $f_2'$ are elementary, and therefore $f'$ is elementary. Similarly for the other possibilities for the last step.

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  • $\begingroup$ Okay then how can we prove the derivative of an elementary function is always an elementary function? $\endgroup$
    – user604496
    Mar 27, 2019 at 18:26
  • $\begingroup$ @TheGreatDuck, before anything else, what would you say about the function $\frac{x+\sqrt{x^2}}{2x}$? $\endgroup$ Mar 28, 2019 at 2:46
  • $\begingroup$ The Heaviside step function is elementary, according to (2), as it satisfies $f^2 - f = 0$. $\endgroup$ Mar 28, 2019 at 3:01
  • $\begingroup$ @RobertIsrael: So then, any $\{0,1\}$-valued function is elementary? Even, say, the indicator function of a non-measurable set? That seems peculiar. $\endgroup$ Mar 28, 2019 at 4:07
  • $\begingroup$ Actually I think I have to take that back. The proper setting for differential algebra is a differential field, i.e. a field equipped with a derivation. The problem with the Heaviside function is not with its derivative (you can define that as $0$), but that it can't be a member of a field: it is not $0$, but you can't divide by it. $\endgroup$ Mar 28, 2019 at 4:45
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No, the derivative of an elementary function is elementary; some integrals were defined specifically as the antiderivative of certain functions because that function otherwise would have no closed-form antiderivative.

An anti-derivative of a non-elementary function cannot be an elementary function.

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Yes, and I can provide a simple counter-example.

Let $f(x)$ be piece-wise defined such that $f(x) = x^2$ for $x \neq 0$ and such that $f(0) = 300$.

This is not an elementary function. However its integral is $F(x) = \frac {1}{3}x^3 + c$ which is elementary.

For a slightly more "non-elementary" example just make $f(x) = -500$ whenever $x$ is an integer multiple of $n = 0.0001$. Feel free to keep decreasing $n$ to make the function messier and messier.

However, if you want a continuous non-elementary $f$ then no. If $f$ is continuous then by one of the fundamental theorems of calculus $F'(x) = f(x)$ and the derivative of an elementary function is an elementary function. Furthermore, if you want that $f$ is an integral of some other $h$ then it follows that $f$ is continuous as the integral of any real valued function defined everywhere is a continuous function. So this will only work with discontinuous $f$'s that are not integrals of other functions.

In short the set of derivatives of elementary functions $\neq$ the set of anti-integrals of elementary functions.

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    $\begingroup$ Why is a piecewise function with elementary cases not elementary? $\endgroup$ Mar 28, 2019 at 1:20
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    $\begingroup$ I am talking about your second sentence. You give a quadratic function with a hole and say that it is nonelementary. $\endgroup$ Mar 28, 2019 at 1:43
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    $\begingroup$ "Elementary has imo always been a subjective concept." - in this regard at least, we are in agreement. $\endgroup$ Mar 28, 2019 at 1:55

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