# When is it impossible to integrate?

1. Feb 16, 2010

### hover

I've heard that there are some types of functions that are impossible to integrate. Can I see an example and why so?

Thanks!

2. Feb 16, 2010

### FallenRGH

It's not that they are impossible to integrate, but rather their anti derivative cannot be defined by an elementary function( for example, the integral of ex2 or the integral of sin(x2). They can however be approximated if they are definite integrals using a Riemann sum, or using more complex methods of integration(for example with sin(x2) using the Fresnel Integral).

3. Feb 16, 2010

### ice109

or he could be talking about functions that are not riemann integrable like the http://en.wikipedia.org/wiki/Indicator_function" [Broken]

Last edited by a moderator: May 4, 2017
4. Feb 18, 2010

### Count Iblis

It is easy to prove that the indicator function of the quotient set [0,1]/Q cannot be integrated over the interval [0,1].

5. Feb 18, 2010

### Anonymous217

It's e^{-x^2}, not e^{x^2}. Just correcting a typo most likely.

6. Feb 19, 2010

### Char. Limit

One way to find a function that isn't integrable is, perhaps, to find a function that is nowhere differentiable. Its derivative (aside from being an impossibility) is not integrable.

7. Feb 19, 2010

### FallenRGH

Ah, yes. Sorry, and thank you.

8. Feb 19, 2010

### hover

Well it makes sense that you can approximate any integral using Riemann sums as long as bounds are provided. That is the definition of an integral.

My calc book says that you can integrate elementary functions. Just out of curiosity, how can you tell when you don't have an elementary function?

9. Feb 19, 2010

### Tac-Tics

Elementary is kind of a fuzzy word. To give one possible definition for it, it would be:

polynomials, trigonometric functions, exponential functions, the logarithm functions, and any finite sums, products, and compositions thereof.

We know the rules to integrate polynomials, trigs, exps, and logs. We also have the product rule, the chain rule, and linearity (meaning (f+g)' = f' + g'), which allows us to handle all the sums, products, and comps.

The way you know something is elementary or not is by proving it.

If I have sin^2(x) + cos^2(x^2), it's elementary because it follows the definition. Sin, cos, and x^2 are elementary, and the above function can pieced together from those three.

If I have e^(-x^2), it's also elementary, because, again, it satisfies the definition.

If I have erf(x), which is the derivative of e^(-x^2), we have a problem. It's not inherently obvious what erf is. Maybe it's something like blah * e^blah or something, but we can't tell by looking at its definition. We need to investigate some of erf's properties. I don't know how you do it, personally, but some guy figured it out that it requires an infinite number of sums, products, and comps to produce erf(x). Therefore, it's non-elementary.

Non-elementary functions can't be "written" down in algebra. They are produced by definitions, usually by some integral or differential equation, not an algebraic equation.

10. Feb 24, 2010

### Gib Z

In general humans are not able to tell. There is something called the Risch algorithm that finds out if elementary antiderivatves exist, but its complex. So much so the full version of the algorithm has never been programmed onto a computer, though it is theoretically possible to do so.

11. Feb 24, 2010

### Tac-Tics

This has nothing to do with being human.

The Risch algorithm is not a well-specified algorithm. It has nothing to do with the complexity.

12. Feb 24, 2010

### Fredrik

Staff Emeritus
The integral is supposed to give you the area under the graph. You can overestimate that area by computing the total area of a finite number of rectangles that stand side by side with the top edge of each rectangle above the graph. You can also underestimate the area by using rectangles with the top edge under the graph. If the greatest lower bound of the set of all such overestimations is different from the the least upper bound of the set of all such underestimations, the function isn't Riemann integrable.

13. Feb 24, 2010

### Werg22

What?

14. Feb 25, 2010

### Char. Limit

Yeah, good luck getting some sense out of that. Looking at it again, it doesn't even quite make sense to me.

15. Feb 25, 2010

### crd

the way i translated it was "if the area of the smallest big rectangle is not equal to the area of the biggest small rectangle, then the function is not Riemann integrable"

16. Feb 26, 2010

### Fredrik

Staff Emeritus
If that's in response to my post, what I said is closer to (but not exactly the same as) "if the smallest sum of areas of big rectangles is not equal to the biggest sum of areas of small rectangles., then the function is not Riemann integrable". The main difference between that and what I said is that the "smallest" member of a set doesn't always exist. For example, what is the smallest member of the set of all x such that 1<x<2? That's why we talk about the greatest lower bound instead. (For the set I used as an example, it's =1). Same thing with "biggest" and "least upper bound".

The precise statement can be found here (and in lots of other places).

17. Feb 26, 2010

### sEsposito

I'm pretty sure that it's going to be hard to actually find a function that is 100% not integrable. The reason being that most functions -- especially those in textbooks -- are contrived from applications in science, engineering etc.

That being said, a reimann sum is considered the safe way to integrate when all other bets are lost. So, that would be what I would consider the ultimate test.

I may run this past one of my professors, as the question is quite an interesting one.

18. Feb 26, 2010

### jgens

It's actually easier than you might think. For example, consider the function $f$ defined by $f(x) = 0$ for rational $x$ and $f(x) = 1$ for irrational $x$. Now, let $[a,b] \subset \mathbb{R}$ and suppose that $P = \{t_0, \dots, t_n\}$ is a partition of $[a,b]$ such that $t_0 = a < t_1 < \dots < t_{n-1} < t_n = b$. Clearly the lower Darboux sum $L(f,P)$ for this function $f$ and partition $P$ is zero (since every interval contains at least one rational number). Additionally, the upper Darboux sum $U(f,P)$ for the function $f$ and partition $P$ is $b - a$ (since every interval contains at least one irrational number). Therefore, $\sup\{L(f,P)\} < \inf\{U(f,P)\}$ so $f$ is not Darboux integrable and consequently, is not Riemann integrable too.

It's actually possible to find even simpler examples of functions which are not integrable, like the function $f:\mathbb{N} \to \mathbb{N}$ defined by $f(n) = n$. However, it's fairly clear from the context that people are (generally) talking about real valued functions of a real variable.

19. Apr 15, 2010

### shree007

the functions like ' sinx/x , tan rootx,