20
$\begingroup$

Say, for example, I take a reasonably-complicated function $f(x)=\tanh[\ln(x^x)]$, and differentiate it to get $$f'(x)=\frac{4x^{2x} [1+\ln(x)]}{(x^{2x}+1)^2}.$$

Now, to integrate this, I imagine, would be very difficult and time-consuming.

My question is: does there exist a function $f$ whose derivative $f'$ we can't integrate (without differentiating $f$ in the first place), using substitution, parts and/or partial fractions (or other integration methods)?

My motivation for this is that it's very easy to differentiate even a ridiculously-complicated function (using the chain and/or product rule), but I've often wondered whether we could get back to the original function by integrating this derivative.

If I'm not articulating myself clearly enough, please ask me to explain further.

Thanks!

Edit

I know, from the Fundamental Theorem of Calculus, that such a function can be integrated, but, other than knowing the fact that this is the derivative of a suitable function, could it be impossible to reverse-engineer the problem, to get $f'$ back to $f$ (using only methods of integration)?

| cite | improve this question | |
$\endgroup$
  • 6
    $\begingroup$ Doesn't the fundamental theorem of calculus settle this ? $\endgroup$ – Tom Collinge Jun 23 '14 at 14:25
  • 2
    $\begingroup$ I remember in Surely You're Joking Mr Feynman, Richard Feynman referred to a real integration problem (with a real solution) but which could only be solved using complex analysis. Would such an example meet your criterion? $\endgroup$ – lemon Jun 23 '14 at 14:26
  • $\begingroup$ @TomCollinge Well, the FTC says that differentiation is the reverse process of integration; I'm talking not talking about whether the integral exists or not: I'm referring to how one could reverse-engineer the integral back into the differential via $u-$substitution, etc. $\endgroup$ – beep-boop Jun 23 '14 at 14:28
  • 1
    $\begingroup$ This MathOverflow question seems relevant. $\endgroup$ – Santiago Canez Jun 23 '14 at 14:45
  • 2
    $\begingroup$ This is not an answer, but in high school I was always amused by the fact that to integrate $\ln(x)$ you have to 'multiply it by $1$' and then integrate by parts (maybe there are other ways too, but this is the one normally taught). If someone doesn't tell you about this, you might spend an eternity trying to figure it out, but once you know the 'trick', it's easy. $\endgroup$ – MGA Jun 23 '14 at 15:03
33
$\begingroup$

The answer is no.

While the methods you learn in calculus like partial fractions, integration by parts, etc. are going to be limited, there is a complete algorithm, called the Risch algorithm for integrating elementary functions. Complete means that for any elementary function, the algorithm will either find an elementary antiderivative, or prove that none exists (note that an antiderivative always exists, but it will not always be an elementary function, for instance, it's well known that $\int e^{-x^2}\, dx$ is not an elementary function).

Since the derivative of an elementary function is always an elementary function, the Risch algorithm applied to any derivative of an elementary function will always work. It may not produce exactly the same function back, since the antiderivative is only defined up to an additive constant (i.e., you might get your original function back plus $C$).

Note that the Risch algorithm itself is very complicated, and involves some deep algebra, and a ton of different technical cases. It's much more complicated than the tricks you learn in calculus.

Many computer algebra systems implement some subset of the Risch algorithm. For instance, Mathematica (or Wolfram Alpha) should be able to integrate just about any derivative of an elementary function that you throw at it.

If you're interested in a high level view of the Risch algorithm, I recommend Manuel Bronstein's Symbolic integration tutorial. If you want a more detailed view (with pseudocode to implement in a compute algebra system), I recommend his book, Symbolic Integration I: Transcendental Functions.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Important point: integration algorithms such as the Risch algorithm require that the field of constants be explicitly computable. Without such, the problem of integration is undecidable in general. And, by old results of Daniel Richardson, most of the constant fields used in calculus are undecidable. So, technically, the answer might be "yes", depending on how the problem is presented. $\endgroup$ – Bill Dubuque Jun 24 '14 at 0:17
  • 1
    $\begingroup$ That's true, but it's not special to the Risch algorithm. Virtually every algorithm in algebra has this restriction. It boils down to simple things like the fact that $ax^n$ is degree $n$ in $x$ only if $a\neq0$. And note how zero equivalence computability in the constant field is fundamental to the correctness of very basic algorithms like the division algorithm and Gaussian elimination. $\endgroup$ – asmeurer Jun 24 '14 at 1:15
  • $\begingroup$ It depends on the algebraic context. If one tries to make effective elementary calculus then one is faced with difficult or undecidable problems in transcenddental number theory when one attempts to calculate effectively with constants like $\,\pi, e,\,$ and other special values of transcendetal functions (not to mention the functions themselves). Theoretically these are real problems, even if some of then pose little difficulty in practice. $\endgroup$ – Bill Dubuque Jun 24 '14 at 2:16
  • $\begingroup$ By the way, only the algebraic case of the Risch algorithm is "very complicated", i.e. integrands involving algebraic functions such as $n$'th roots, etc. The purely transcendental case can easily be comprehended by anyone who understands a little ring theory. I did the implementation of a variant of Bronstein-Risch for Macsyma, so I speak from firsthand experience. $\endgroup$ – Bill Dubuque Jun 24 '14 at 2:26
  • $\begingroup$ Maybe complex is a better word than complicated. $\endgroup$ – asmeurer Jun 24 '14 at 15:42
8
$\begingroup$

The question as posed has no answer. As opposed to differentiation, integration is not a well-defined operation: there is not a fixed set of rules to apply (as you have with differentiation of elementary functions). So the question of whether an anti-derivative cannot be found depends on a person's ability/ingenuity.

That anti-derivative that looks impossible to you, might be obvious to someone else.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ @alexqwx For instance, to anyone who has seen that exact derivative before, the integration is trivial (sometimes this is called "integration by recognition"). $\endgroup$ – MGA Jun 23 '14 at 14:59
5
$\begingroup$

The fundamental theorem of calculus tells us that if $F$ is any antiderivative of $f'$ then $$\int_{a}^{x} f'(t)dt = F(x) - F(a)$$ since $f$ is already an antiderivative of $f'$ we have: $$\int_{a}^{x} f'(t)dt = f(x) - f(a)$$ and we have our integral.

I should add that for many of functions we ``know" how to integrate, the antiderivative was found by simply by taking a derivative and not necessarily by using techniques like u-substitution.

For instance, historically speaking the antiderivative of $1/x$ would fit your question. For a long time people were trying to determine the area bounded by this curve. It was the question of quadrature of the hyperbola by Archimedes. It wasn't until they realized that the areas behaved somewhat like a logarithm that any progress could be made.

| cite | improve this answer | |
$\endgroup$
  • 9
    $\begingroup$ This is not what the OP is asking. His question is, ok, we know that since $(-\ln (\cos x))^\prime = \tan x$ then $\int \tan x dx = -\ln(\cos x)$ but, would you know it without actually knowing the derivative above? Is there any function such that, once derivated, taking the integral of the derivative renders a calculation that cannot be approached by any reasonable method besides being the derivative of a known function? I myself find it a very good question. $\endgroup$ – busman Jun 23 '14 at 14:31
  • $\begingroup$ I am not saying it's a bad question. But generally speaking, if something is the derivative of a function, there is some way of going backwards, even if it is just searching for the right function until you find it. $\endgroup$ – Joel Jun 23 '14 at 14:36
  • $\begingroup$ @busman Yes! That's precisely what I'm asking! $\endgroup$ – beep-boop Jun 23 '14 at 14:36
  • $\begingroup$ $\int \tan(x)\,dx$ is not an example of this - it is easily evaluated by substitution. $\endgroup$ – lemon Jun 23 '14 at 14:38
  • $\begingroup$ @user2584283 Yes, but you get the gist of what they mean. $\endgroup$ – beep-boop Jun 23 '14 at 14:39
1
$\begingroup$

Sorry if I am misunderstanding your question. But let me give you a fact(which may not be related to this question): Here is example of function which is bounded, derivative of some function but not Riemann Integrable: http://en.wikipedia.org/wiki/Volterra%27s_function

FTOIC applicable when derivative of function is integrable. It is not always applicable

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

The fundamental question is: "Given a function, does its integral exist?" If the function F' is created by taking the derivative of F, then, obviously, the integral of F' MUST exist, as F has already been proven to exist.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm not disputing its existence; $\int f'(x)dx$ exists trivially, by FTC, since we've computed $f'(x)=\frac{df}{dx}$ . I'm asking whether there is a function whose integral we can find only by knowing that it is the derivative of some other function. $\endgroup$ – beep-boop Jun 23 '14 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.