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I Indefinite integral

  1. Sep 24, 2017 #1
    I was wandering if there is a way to understand whether it is possible to find an indefinite integral of a function. Let's say e^(-x^2) or e^(x^2)... They can't have indefinite integrals, but how can I say it? Is there a theorem or something?
     
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  3. Sep 24, 2017 #2

    andrewkirk

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    They do have indefinite integrals, it's just that those indefinite integrals cannot be written using the tools that are most commonly used to define functions, which are:
    • arithmetic functions ##\times,+,-,\div##
    • exponential, log and trig functions
    • parentheses
    • cases (eg ##f(x)=0## if ##x<0## otherwise ##1##)
    After choosing a set of tools like this, call it S, we can recursively define a class E of functions that can be defined in a finite sequence of symbols using tools from S. then the statement you want to make about f not having an indefinite integral is that 'there is no function in E whose derivative is f'.

    Note that the class E varies according to the tools we allow. The function ##f(x)=e^{-x^2/2}## has no indefinite integral in the class E based on tools in the four bullet points above, but if we add the function ##\Phi## which is used to denote the cumulative distribution function of the standard normal distribution, to the toolbox, then ##f## does have an indefinite integral, which is the set of functions ##\{g:\mathbb R\to\mathbb R\ :\ g(x)=\Phi(x)\sqrt{2\pi}+C,\ C\in\mathbb R\}##.

    The gamma function ##\Gamma## is another example of a tool that might be added. IIRC it is an indefinite integral that is not expressible using the tools in the four bullets above. Various Bessel functions might be other examples.

    For any class of functions E and function f such that there is no indefinite integral of f in E, we can define a new class E* that is constructed from the toolbox of E together with the indefinite integral of f.
     
  4. Sep 25, 2017 #3
    Thank you! Very clear! So, now my question is: "how can I understand if f has a primitive in E?" Is it possible to see it without trying to solve the integral?
     
  5. Sep 25, 2017 #4

    andrewkirk

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    In general it's not possible. The only way to know for sure in an arbitrary case is to search all the functions in E. But since E is an infinite set, it cannot be searched in finite time. In certain specific cases there may be specific proofs that E does not contain a primitive. But those proofs will not be generalisable.
     
  6. Sep 25, 2017 #5
    So, how can they say e^-(x^2) has not primitive in E? The only thing that can be said is that the primitive has not yet been found, right?
    PS: how can I write it with LaTex? I can't find a way...
     
  7. Sep 25, 2017 #6

    andrewkirk

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    It may be that for that particular function there is a specific proof of the lack of a primitive. But I don't think the method used will be generalisable to an arbitrary function.

    Or, it may be that no proof has been found and the author was just being a bit loose with their words.

    If you know Latex, the only site-specific thing you need to know to make it work here is that the delimiter for in-line latex is ## rather than $. The delimiters for display latex are the same.
     
  8. Sep 25, 2017 #7

    fresh_42

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