Register to reply

Functions that cannot be integrated

by trdiayw
Tags: functions, integrated
Share this thread:
trdiayw
#1
Nov8-06, 07:42 AM
P: 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?

Thanks,

Trdiayw
Phys.Org News Partner Science news on Phys.org
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
HallsofIvy
#2
Nov8-06, 08:25 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310
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.
mathwonk
#3
Nov8-06, 08:51 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,453
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".

Sane
#4
Nov8-06, 09:35 PM
P: 223
Functions that cannot be integrated

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.

[tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]

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.
VietDao29
#5
Nov9-06, 05:58 AM
HW Helper
VietDao29's Avatar
P: 1,421
Quote Quote by Sane
[tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]

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.
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)
Sane
#6
Nov9-06, 08:00 AM
P: 223
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...
leon1127
#7
Nov12-06, 11:59 PM
P: 487
Quote Quote by Sane
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...
it is proven that only certain fucntions have elementary antiderivative.
http://www.claymath.org/programs/out...s05/Conrad.pdf

if we know they have elementary antiderivative, we just need to play with trig, inverse trig, log, exp, polynomial, and intergration by parts.
ObsessiveMathsFreak
#8
Nov13-06, 03:45 AM
P: 406
[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 .
Office_Shredder
#9
Nov13-06, 04:07 AM
Emeritus
Sci Advisor
PF Gold
P: 4,500
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
George Jones
#10
Nov13-06, 05:58 AM
Mentor
George Jones's Avatar
P: 6,226
Quote Quote by ObsessiveMathsFreak
[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 .
How is this different from

[tex]\int \mathrm{sin} x dx? [/tex]
Tleilaxu_Ghola
#11
Nov27-06, 12:59 PM
P: 4
Quote Quote by ObsessiveMathsFreak View Post
[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 .
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.
Gib Z
#12
Nov28-06, 04:42 AM
HW Helper
Gib Z's Avatar
P: 3,352
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).
HallsofIvy
#13
Nov28-06, 06:14 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310
Quote Quote by Office_Shredder View Post
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
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!
Karthikthe
#14
Jun19-09, 01:15 AM
P: 5
i think u can integrate the function by expanding it over infinite terms
HallsofIvy
#15
Jun19-09, 06:44 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310
Quote Quote by Karthikthe View Post
i think u can integrate the function by expanding it over infinite terms
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.
HallsofIvy
#16
Jun19-09, 06:50 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310
Quote Quote by Gib Z View Post
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).
[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.
Jinius
#17
Jan30-10, 03:27 AM
P: 2
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
wofsy
#18
Jan30-10, 06:46 AM
P: 707
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.


Register to reply

Related Discussions
Integrated MSc in physics Academic Guidance 26
Woohoo I have integrated Lnx General Discussion 0
The integrated background intensity? Advanced Physics Homework 0
Choosing what's integrated Calculus 2
Integrated rate law Biology, Chemistry & Other Homework 2