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Since the integration of a function is the opposite of a the derivative of a function, and there are clear steps to follow when we want to find the derivative of a function, I thought there would be clear steps to follow too when we want to find the integral of a function.

What is it about finding the integration of a function that makes it impossible to come up with a guaranteed method to find it?

Edit: I am only considering elementary functions.

mauna
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Because the class of function you consider easy to differentiate—namely, the elementary functions—is not mapped onto itself. You are accustomed to thinking of elementary functions as being exactly those on which you "do calculus" because they are easy to differentiate, but a randomly chosen one may not be in the image of the derivative function, and therefore, not easy to integrate. The standard example $f(x) = e^{x^2}$ applies here; no simple example of a hard-to-differentiate function exists, by definition.

Note that without the assurance that functions are given by explicit formulas of a convenient form, neither integration nor differentiation is particularly easy.

Ryan Reich
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There are clear steps to follow when we want to compute the derivative of a function, only for a narrow class of functions. In general no algorithm is known for differentiation. The same holds for integrals.

EDIT: If in the search for further information, turn to the link, kindly suggested by Chris Janjigian.

Community
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superAnnoyingUser
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    It would be nice to add an example of such a "no clear steps to find the derivative" function. – Xoff Dec 09 '13 at 12:37
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    That would be really hard, as all elementary functions have a well known derivative, @Xoff. – superAnnoyingUser Dec 09 '13 at 12:46
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    If you say "in general", then it must be easy to find an example. If not, then don't say "in general". – Xoff Dec 09 '13 at 12:56
  • So your answer to "why isn't there a fixed procedure" is "because there is no fixed procedure"? – Najib Idrissi Dec 09 '13 at 12:58
  • @Xoff I disagree. If most (in any sensible sense) functions don't admit easily computable derivatives, then "in general" is proper even if an example cannot be easily shown (of course, we could point to the 2nd antiderivative of $x^x$ in $(0,1)$, except that a derivative can be ascertained in those same terms). – Jonathan Y. Dec 09 '13 at 13:30
  • @nik, as I see it, the answer is "there's no fixed procedure with differentiation either", which is not quite the same thing. – Jonathan Y. Dec 09 '13 at 13:31
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    @JonathanY. I understand your point of view, because you think about the set of functions, and in general means "for almost all". But for the OP, I'm pretty sure it means "for almost all functions I will encounter", and it's not the same thing. If you can't give an example of something and you claim that this thing is almost always occurring, it's a real pedagogical problem. – Xoff Dec 09 '13 at 13:37
  • @Xoff I agree that this is a problem, but I can't see that the OP is well-served by sweeping it under the carpet. Perhaps something of the sort of the accepted answer on the Math Overflow question RahulNarain linked to above is suitable here; I just meant that any answer should reflect that the apparent asymmetry is due to our own very limited system of notation. In any case, downvoting this answer strikes me as disingenuous (not implying that you did that; in fact I quite suspect someone else is responsible). – Jonathan Y. Dec 09 '13 at 14:41
  • Your comments are not in general constructive and not even in general true, @Xoff. To be more specific, saying something is true in general, does not mean it is easy to find an example, as in this case. Yet what I said in my answer is true in general, hence the use of the expression. – superAnnoyingUser Dec 09 '13 at 19:21
  • I didn't want to be rude. It's just that when I teach my students, if I say something like that, they would not be convinced. That's why I wrote those comments. – Xoff Dec 09 '13 at 20:12
  • Well thank you very much, sir. I've been thinking about such an example since your first comment. – superAnnoyingUser Dec 09 '13 at 20:14
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For special types of functions there are pretty clear cut methods for integrating (e.g. polynomials). Unfortunately, there is no general algorithm for creating integrals as there are functions that have no antiderivative. The classic example being $\ f(x)=e^{x^2}$.

user114603
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There is an algorithm for computing indefinite integrals in terms of elementary functions, when such representations exist: Risch's algorithm . It is a very complex algorithm which is now included in the major symbolic mathematics programs.

superAnnoyingUser
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lhf
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In order to answer your question, one has to clarify it. You presumably mean integrating explicitly elementary functions in terms of further elementary functions. Of course, a complete answer would required a precise definition of this term but we can regard it informally as meaning one which is obtained from the usual suspects (powers, trigonometric, exponential, logarithm, etc.) by simple operations (arithmetic, composition, etc.). The difference is then that while there are simple rules which describe what happens when differentiating, there are no such ones for integration. For example, integration by parts merely shifts the problem to a new integral---this might work in a particular case but not in general. You can see this effect when you use Mathematica which cannot integrate all elementary functions explicitly. By the way there is, rather surprisingly, a general theory which covers this topic and this is used in the algorithm employed by Mathematica.

mathuser5891
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  • I see that the algorithm I mentioned is pinpointed in the answer below of Ihf which was posted while I was at the coal face. – mathuser5891 Dec 09 '13 at 12:51