# Limits of integrals

1. Sep 26, 2006

### gonzo

What are the rules for when you can and can not move the limit of a definite integral inside the integration sign?

For example, if I have a definite integral of a function f(z) and the function includes a constant k, and I want to take the limite of the definite integral as k goes somewhere interesting, under what circumstances can I take the limit first of f(z) before interating and get the same answer (if ever)?

Note that this limit has no affect on the limits of integration.

For example:

$$\lim_{r\to\infty} \int_{0}^{2\pi}{e^{r\alpha}e^{ix\alpha} \over 1+e^re^{ix}}dx$$

for $0<\alpha<1$

If I can move the limit inside the integral, it is is easy to show that this is 0. Otherwise it is more annoying (and I am pretty the answer here has to be zero).

2. Sep 26, 2006

### shmoe

You've probably seen swapping the order of a limit and an integral in the real case? You can apply the relevant theorems, uniform convergence will do the trick (for your example).

Alternatively for this example, you can try to bound the absolute value of that integral by something that tends to 0 as r->infinity (you'd do this anyways trying to prove uniform convergence).

3. Sep 26, 2006

### gonzo

Actually, that's part of the problem, I haven't seen this in the real case, or at least I don't remember it.

Can you point me to the relevant theorems? Or at least give me a quick overview or summary?

4. Sep 26, 2006

### gonzo

To be clear, I understand the idea of uniform convergence in general, and I sort of understand the second thing you said for a way to show it in this case, but I was wondering about general theorems, which I haven't seen. And in case it wasn't clear in this example, x itself is real.

5. Sep 26, 2006

### shmoe

This will be in any real analysis text, you've probably seen integrating power series term by term? It's a consequence of uniform convergence, and a more general theorem is usually proven first.

If you have a sequence $$f_n$$ of integrable (real valued) functions on [a,b] that converge uniformly to f on [a,b] then f is integrable and you have

$$\lim\limits_{n\rightarrow\infty}\int_a^b f_n(x)dx=\int_a^bf(x)dx$$

An integral of a complex function can be viewed as the integral of the real part plus the integral of the imaginary part and you can apply the above.

If you want to talk about integrals on infinite intervals then uniform convergence won't be enough. there are various criterea that allow interchanging the limit and the integral, see the Lebesgue dominated convergence theorem for example.

6. Sep 26, 2006

### gonzo

I understand what you wrote above, but I don't really see how it applies to my issue. Your limit example is with regard to the number of terms in a series, which seems to me to be different, even if you talk about expanding a funciton with a power series.

I'm also sure I can probably solve my particular by a more detailed analysis, but I was hoping that there would be some useful theorems about when specifically you can move the limite inside the integral, since that would greatly simplify things.

I assume this is dependent on some quality of the function being integrated, since a casual look shows that this works for simple functions.

I would like to know if there is a simple set of conditions on a function being integrated that are sufficient for this to be true.

7. Sep 26, 2006

### shmoe

If you can prove the version I gave for sequences of functions, you should see the result is true for a family of functions f_r(x) converging to f(x) uniformly in r, a real variable.

to prove the result in my last post (a simpler version, let's assume we know f is integrable), for any $$\epsilon>0$$ you can find an integer N such that n>=N implies $$|f_n(x)-f(x)|<\epsilon/(b-a)$$ for all x in [a,b]. Then:

$$\left|\int_a^b f_n(x)dx-\int_a^bf(x)dx\right|\leq\int_a^b|f_n(x)-f(x)|dx\leq\epsilon$$

So the limit of the integrals converges to the integral of the limits.

If you have a family of functions parameterized by real numbers r that converges uniformly, just change all the n's to r's and N's to R's and the proof works just fine.

8. Sep 26, 2006

### gonzo

Okay, let me see if I'm following everything. We'll use my example.

Let:
$$f={e^{r\alpha}e^{ix\alpha} \over 1+e^re^{ix}}$$

Then if we can show that f uniformly converges to 0 as r goes to infinity over the domain of integration then we can move the limit inside the integral, which in this case shows that the integral also goes to 0.

Since we are working with complex number, we want to show that the absolute value of f converges uniformly to 0.

This gives us:

$$|f|=\sqrt {e^{r\alpha} \over {1+2e^rcosx+e^{2r}}}$$

And then we have:

$$|f| \leq \sqrt {e^{r\alpha} \over {1-2e^r+e^{2r}}}$$

This last expression is independent of x, and converges uniformly to 0, so the absolute value of f must also do so, which means our integral is 0.

Is that about what you meant?

9. Sep 26, 2006

### shmoe

Your numerator shouldn't be squared, otherwise that's fine.

You don't have to bother multiplying by the conjugate though, the absolute value of the numerator is just e^(alpha*r). The denominator's absolute value is >= e^r-1 by the triangle inequality, that same as what you have but less big square root signs.

10. Sep 27, 2006

### gonzo

Sure, that was sloppy of me, but I was more concerned with getting the general idea correct. Which from your response I assume it is.