Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functions that cannot be integrated

  1. Nov 8, 2006 #1
    Hi all,

    I am in Calculus II now, and after studying several techniques to integrate functions I started wondering about functions that either cannot be integrated or are so time consuming and complex to integrate that it becomes impractical.

    How do we study such functions? Can we still gather information about the function?


  2. jcsd
  3. Nov 8, 2006 #2


    User Avatar
    Science Advisor

    If you are referring to functions that, theoretically, have an integral but it's not in any simple form, there are two options: approximate the function by something you can handle or do a numerical integration.

    I'm not sure what you mean by "gather information". The integral seldom gives a much information about a function as the derivative.
  4. Nov 8, 2006 #3


    User Avatar
    Science Advisor
    Homework Helper

    what is your definition of a function that "can be integrated"?

    e.g. any bounded function on a bounded interval, whose discontinuities can be overed by a sequence of intervals of toital length less than any arbitrarily given positive number, has a riemann integral.

    are you talking about the problem of finding "elementary" antiderivatives of "elementary" functions? if so you need to define "elementary".
  5. Nov 8, 2006 #4
    Any function that is differentiable between a and b can be integrated between a and b by taking the area under the curve. This area can be approximated by calculating the riemann sum of an infinite number of rectangles, using a limit. If you're suggesting the possibility of attaining the integrated equation in exact form, I can't help you out there. There are advanced integration methods that cover most situations that you would ever encounter, as far as I know. But there must be some that are inplausable to calculate.


    For example. The integral of that is an inverse sine wave. However, I'm not sure if you can solve that without using the process that was used to determine the derivative of an inverse sine wave. In a different situation, you wouldn't know the derivative of the function, in order to determine the proof for finding the integral of the derivative. I could guess that this makes some equations "impossible" to prove.
    Last edited: Nov 8, 2006
  6. Nov 9, 2006 #5


    User Avatar
    Homework Helper

    I can solve this integral by letting x = sin(t), with [tex]t \in \left[ - \frac{\pi}{2} ; \ \frac{\pi}{2} \right][/tex]
    [tex]\int \frac{dx}{\sqrt{1 - x ^ 2}} = \int \frac{\cos t dt}{\sqrt{1 - \sin ^ 2 t}} = \int \frac{\cos t dt}{|\cos t|}} = \int \frac{\cos t dt}{\cos t}[/tex]
    Since cos(t) is positive for: [tex]t \in \left[ - \frac{\pi}{2} ; \ \frac{\pi}{2} \right][/tex]
    [tex]\int \frac{dx}{\sqrt{1 - x ^ 2}} = ... = \int dt = t + C = \arcsin (x) + C[/tex] (Q.E.D)
  7. Nov 9, 2006 #6
    I think you missed my point. I was saying you couldn't solve that if you didn't know its origins were a trigonmetric wave in the first place...
  8. Nov 12, 2006 #7
    it is proven that only certain fucntions have elementary antiderivative.
    http://www.claymath.org/programs/outreach/academy/LectureNotes05/Conrad.pdf" [Broken]

    if we know they have elementary antiderivative, we just need to play with trig, inverse trig, log, exp, polynomial, and intergration by parts.
    Last edited by a moderator: May 2, 2017
  9. Nov 13, 2006 #8
    [tex]\int e^{-x^2} dx[/tex]

    This function is essentially the be all and end all of "non-integrable" functions. Though at this point, the numerical integral of this function can be computed in many cases faster than a good number of "known" functions .
  10. Nov 13, 2006 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Even if you have something that's not "integrable", as long as it's differentiable you can find the taylor's series and integrate it term by term. It's not pretty, but it works
  11. Nov 13, 2006 #10

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How is this different from

    [tex]\int \mathrm{sin} x dx? [/tex]
  12. Nov 27, 2006 #11
    Isn't that integral really similar to the normal distribution?

    Substitute x^2 for k^2/2 => x = 2^-.5*k

    Then you have the normal distribution without the usual normalization constant of 1/(2 pi)^.5,

    So that integral should be equal to... pi^-.5. Assuming I did my math correctly. (if the integral is taken from negative infinity to infinity). To find smaller, arbitrary intervals you could use all the familiar stats tricks with normal densities.

    IIRC, the way to actually calculate that out was to expand it to a double integral and then use a trig-substitution.
  13. Nov 28, 2006 #12

    Gib Z

    User Avatar
    Homework Helper

    Just a side note:

    [tex]\int e^{-x^2} dx[/tex]

    when integrated from +infinity to -infinity gives you Root Pi, if my memory serves me well. O and yes, it is quite a good approximation to the normal distribution, as is 1/(1+x^2).
    Last edited: Nov 28, 2006
  14. Nov 28, 2006 #13


    User Avatar
    Science Advisor

    As long as it is infinitely differentiable, you can find its Taylor's series. Of course, it might not be equal to that Taylor's series so the integral you get might not be correct. In order to find an integral as an infinite series by integrating the Taylor's series term by term, you would have to have an analytic function.

    Since analytic functions are very "nice" and the only requirement for integrability is that the function be bounded and have discontinuities only on a set of measure 0, jumping form integrable to analytic leaves out "almost all" integrable functions!
  15. Jun 19, 2009 #14
    i think u can integrate the function by expanding it over infinite terms
  16. Jun 19, 2009 #15


    User Avatar
    Science Advisor

    What function are you talking about? And what do you mean by "expanding it over infinite terms"? Use the Taylor's series? That had already been said.
  17. Jun 19, 2009 #16


    User Avatar
    Science Advisor

    [tex]f(x)= \frac{1}{2\sqrt{\pi}}e^{-x^2}[/tex]
    is more than a "good approximation"! It is the normal distribution. The integral you gives, from [itex]-\infty[/itex] to [itex]\infty[/itex] is [itex]2\sqrt{\pi}[/itex], making the "area under the curve" for the normal distribution 1 as it should be. The integral from 0 to [itex]\infty[/itex] is [itex]\sqrt{\pi}[/itex].

    And [itex]1/(1+x^2)[/itex] is a "not so good" approximation.
  18. Jan 30, 2010 #17
    I feel any function which shows you a trigonometric function on the numerator and something like x or x^2 in the denominator is non-integrable.( Hey, I know special cases may get them canceled). I am saying this only coz there happen to be infinite terms in expansion of a negative or fractional index
  19. Jan 30, 2010 #18
    On a bounded interval, any function whose discontinuities have measure zero can be Riemann integrated but there is no general technique. Take for instance, the characteristic function of the Cantor set on the unit interval or a 1 dimensional continuous Brownian path on the unit interval.

    On an unbounded domain an integral may be infinite but this really is still integrable.

    Some function's integrals however do not have well defined limits on unbounded domains. These are non-integrable.

    Functions that have discontinuities of positive measure can not be Riemann integrated, for instance the characteristic function of the rational numbers on the unit interval. However, a more general notion of integral, the Lebesque integral, includes this function and many other that can not be Riemann integrated.

    Much of mathematics is figuring out ways to integrate functions.
  20. Jan 30, 2010 #19
    Is there an algorithm that takes as input a function f(x) and numbers a and b (they can be infinite) that can decide if

    [tex]\int_{a}^{b}f(x) dx[/tex]

    has a closed form expression? So, if you take f(x)=exp(-x^2) and a = 0 and b = 1, the algorithm would return: "no", while if you take b to be infinity, it would say: "yes". Of course, the algorithm depends on a precise definition of "closed form expression". If you allow expressions involving the error function, then it would return a different output in this case.
  21. Jan 30, 2010 #20
    I think that you have to be precise about what "closed form expression means". One can always define a new function as an integral and give it a name. Then does that integral suddenly have a closed form solution?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook