Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My calculus teacher told our class that in integral calculus, they teach you how to integrate all kinds of functions by various methods, but in the end tell you that there are infinitely many integrals that we do not know how to integrate (represent in terms of elementary functions).

So, does that mean:

  1. that there exists elementary functions that are unknown in mathematics that can represent the integral,
  2. that the integral can be represented in non-elementary functions (I do not know what those would be...), or
  3. it is absolutely impossible to represent the integral in another way, other than another integral?
share|improve this question
    
I checked questions that came close to mine that could be duplicates, but found none that answered my question... –  zerosofthezeta Aug 5 '13 at 4:15
3  
The third is what is meant (sort of). A good example is the antiderivative of $\exp(x^2)$. We can represent a wealth of functions with power series and we can integrate these power series term by term as we are used to. However, it may turn out that these power series do not agree with any elementary function (or special function) we are familiar with. –  Cameron Williams Aug 5 '13 at 4:17
    
When you attempt to integrate "most" functions the result is not a "finite mix" of functions you see in a calculus book. –  Maesumi Aug 5 '13 at 4:43

3 Answers 3

up vote 3 down vote accepted

There are an infinite number of functions out there, and you can put an integral sign in front of any of them. Some of those functions are pretty strange and/or ugly. There is no reason why such an integral should have a representation in the very limited set of elementary functions. In fact when you get such a representation, you could consider yourself lucky.

For practical purposes, any convergent integral can be evaluated to any desired degree of accuracy. There are numerous numeric methods available with which to do that. And even if an integral should have some elementary function representation, that most often is a lot harder to evaluate than just running a numeric method on the integral. And you wouldn't be more accurate if the answer were cos(28.34) and you had to evaluate that.

Similarly, if you are going to use the integral in a further computation, it might be easiest to leave it as an integral, rather than tangling yourself up in things that have a lot of terms like sec(log^{-1}(x^{2})) -- or much worse.

So why do calculus classes teach you to find anti-derivatives? First, if the problem is going to resolve itself into an easy anti-derivative you might as well use it. Second, to familiarize you better with the underlying concepts, such as the chain rule and the product rule of derivatives. Third to give you some experience with actual answers so you have some idea what the answer ought to look like. If your numeric method evaluates to 2034.86, and you know the answer can't be larger than 80, then you know you made a mistake in your computation.

Of course, nowadays, all those integrals can be accurately evaluated online; but still, you should have some knowledge of what kind of answer to expect and what it all means.

share|improve this answer

Option 2 is probably closer to the truth. As I understand it, this is related to a similar idea from the study of differential equations, which do not always have solutions that can be written in terms of elementary functions. Try reading about "special functions". Examples include the Airy function and the Gamma function.

share|improve this answer

The answer to your question is (3): it is impossible to write the integral in terms of elementary functions (usually functions like $x^n$ for any real $n$, rational, trigonometric, exponential and logarithmic functions) and this fact has been proved for some integrals (so its not (1) which is just a very hard integral). There is not much difference than (2) and (3) because you can just define a non-elementary function as the integral itself.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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