# A continuous functions you can't integrate?

1. Dec 8, 2006

### Alkatran

After reading http://en.wikipedia.org/wiki/Weierstrass_function it occurred to me that I could do the same thing to an integral:

$$\int \sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i} dx$$

$$= \sum_{i=0}^\infty \int \frac{sin(\frac{x}{3^i})}{2^i} dx$$

$$= \sum_{i=0}^\infty -cos(\frac{x}{3^i})(\frac{3}{2})^i$$

which obviously diverges.

If we turn this into a definite integral and take the limit as n goes to infinity, it becomes a conditionally convergent sum. So we have the odd property that the indefinite integral doesn't exist, but the definite one does.

$$\lim_{n \rightarrow \infty} \int_a^b \sum_{i=0}^n \frac{sin(\frac{x}{3^i})}{2^i} dx$$

$$= \lim_{n \rightarrow \infty} \sum_{i=0}^n (cos(\frac{a}{3^i}) - cos(\frac{b}{3^i}))(\frac{3}{2})^i dx$$

$$= \lim_{n \rightarrow \infty} \sum_{i=0}^n -2sin(\frac{a+b}{2*3^i})sin(\frac{a - b}{2*3^i})(\frac{3}{2})^i dx$$
which should converge because sin(x) becomes linear as x approaches 0.

What exactly is the explanation for this? I thought all continuous functions could be integrated (they definitely told us that!).

Last edited: Dec 8, 2006
2. Dec 8, 2006

### StatusX

Indefinite integrals are just definite integrals where one of the bounds, say the upper bound, is a variable. We can usually just ignore the lower bound completely, since starting at a different point just amounts to adding an overall constant, which we don't care about. This would even be fine when you have a finite sum of integrals, ie, ignoring the bounds would be the same as starting each integral at possibly different lower bounds, which isn't strictly correct, but just adds to the final answer a finite sum of constants. But in the case of an infinite sum of integrals, you are shifting the answer by a constant for each term, and the sum of these constant turns out in this case to diverge. Note you can just replace b by x in the second part, change it by a constant if you want, and you have the indefinite integral.

Last edited: Dec 8, 2006
3. Dec 8, 2006

### matt grime

What justification do you have for supposing that this swapping of summation and integral is at all valid? Clearly, in general it is not a valid operation.

4. Dec 8, 2006

### Alkatran

I didn't know that. What kind of situations is it not valid in, besides this one? What if took the limit as n goes to infinity, could I switch it then?

5. Dec 8, 2006

### matt grime

Erm, that infinite sum is *by definition* the limit as 'n tends to infinity'.

It is the standard difference between uniform and conditional convergence.

6. Dec 8, 2006

### StatusX

The series converges absolutely for all x, so I think exchaning the integral and sum as in the second half of post 1 is valid.

7. Dec 9, 2006

### matt grime

I can't say I checked the post very carefully. There are two possible errors. One, is that swapping summation and integration needs justification. The second is: when you did that indefinite integral, each integral comes with a 'plus constant'. Let's call integral F(x). What you appear to be saying is that if we work out F(b)-F(a) by taking limits, then for any a,b this gives a finite answer. But if I just work out F(x) for any x by taking limits I get infinity. That is not very surprising: it is like splitting the sum in the last but one line of maths in your post into two sums. In general you cannot swap the order of summation on things.

Last edited: Dec 9, 2006
8. Dec 9, 2006

### Gib Z

I'm not sure, but i've heard that some functions do not have indefinite integrals, and can only be approximated. Or maybe the authors meant no integral in elementary functions..If true Im thinking fractals...Think of a Koch Snowflake, the definite integral converges, but can not be differentiated, and therefore not integrated, anywhere. Am i correct?

9. Dec 9, 2006

### Alkatran

No, you can integrate some functions that you can't differentiate. See the function in the wikipedia article in my first post: it can be integrated, but not differentiated.

10. Dec 13, 2006

### Alkatran

That's a different situation. For example, the integral of e to the x squared can't be expressed as an elementary function, even though e^(x^2) is an elementary function. You can, however, express it as an infinite sum (I think), but that sum converges absolutely. In this case the sum converges conditionally.

See this wikipedia article: http://en.wikipedia.org/wiki/Elementary_function

11. Dec 13, 2006

### Mute

Does it converge uniformly, though?

You can say it converges along the real axis because

$$\sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i}$$

is clearly bounded by

$$\sum_{i=0}^\infty \frac{1}{2^i}$$,

which converges to 2.

However, if you consider

$$\sum_{i=0}^\infty \frac{sin(\frac{z}{3^i})}{2^i}$$

with z complex, then you can't bound the series because sin(z) grows arbitrarily large along the imaginary axis.

So, maybe I'm missing something , but it looks to me like this series doesn't converge uniformly on the entire complex plane, which would prohibit the swapping of the integral and summation. (Of course, I could be wrong here, given that Fourier series like to converge and I could probably make the same argument there, but we spent a whole semester calling those uniformly convergent in physics, so I may very well be missing something here.)

Last edited: Dec 13, 2006
12. Dec 13, 2006

### StatusX

By the dominated convergence theorem, you can swap the integral and sum (or more specifically, the integral and the limit of the finite sums) if there is an integrable nonnegative function that bounds each of the partial sums. Since:

$$\sum_{i=0}^N \left|\frac{sin(\frac{x}{3^i})}{2^i}\right| \leq \sum_{i=0}^N \frac{1}{2^i}< 2$$

2 is such a function.

Last edited: Dec 13, 2006
13. Dec 14, 2006

### matt grime

I think the assumption, Mute, was that we were only considering this as a function of a real variable.

14. Dec 14, 2006

### Mute

Yes, I realized that, but I was thinking along the lines of how the series for 1/(1+x^2) didn't converge despite having no singularities on the Real axis, but due to the fact it had one on the imaginary axis. I reasoned that perhaps this series might not be uniformly convergent on the complex plane which might limit operations we can do on it with just real calculus. I wasn't sure, though, as I sort of stated at the end, but I figured I might as say something, because at least if I was wrong, someone would correct me and I might learn something. =P

15. Dec 27, 2006