integrating infinite sums


by joshmccraney
Tags: infinite, integrating, sums
joshmccraney
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#1
Feb25-13, 07:16 PM
P: 229
when using the reimann integral over infinite sums, when is it justifiable to interchange the integral and the sum?

[tex]\int\displaystyle\sum_{i=1}^{\infty} f_i(x)dx=\displaystyle\sum_{i=1}^{\infty} \int f_i(x)dx[/tex]

thanks ahead for the help!
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the_wolfman
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#2
Feb25-13, 07:40 PM
P: 135
Always.

∫ a+ b dx= ∫ a dx+ ∫ b dx
Mute
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#3
Feb25-13, 08:01 PM
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Quote Quote by the_wolfman View Post
Always.

∫ a+ b dx= ∫ a dx+ ∫ b dx
No, the sums in joshmccraney's post are infinite sums, so you cannot guarantee that exchanging the integrals and sums is valid. Usually the infinite sum needs to be uniformly convergent to swap the sum and integral.

pwsnafu
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#4
Feb25-13, 09:03 PM
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integrating infinite sums


Quote Quote by Mute View Post
Usually the infinite sum needs to be uniformly convergent to swap the sum and integral.
You also need the integral to be proper. It won't hold on (-∞,∞). Wikipedia gives a counterexample.
jbunniii
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Feb25-13, 09:18 PM
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Quote Quote by the_wolfman View Post
Always.

∫ a+ b dx= ∫ a dx+ ∫ b dx
Here is a counterexample showing that this may not work for infinite sums:

If we define ##g_n(x)## to be a triangular shaped function with base ##0 \leq x \leq 1/n## and height ##2n^2##, then ##\int_{\mathbb{R}}g_n(x) dx = \frac{1}{2}(\textrm{base})(\textrm{height}) = n##. However, ##\lim_{n \rightarrow \infty} g_n(x) = 0## for all ##x##.

Now put ##f_1(x) = g_1(x)##, and ##f_n(x) = g_n(x) - g_{n-1}(x)## for ##n > 1##. We have ##\sum_{n=1}^{N} f_n(x) = g_N(x)##, and so ##\sum_{n=1}^{\infty} f_n(x) = \lim_{N\rightarrow \infty}g_N(x) = 0##. Therefore,
$$\int_{\mathbb{R}} \sum_{n=1}^{\infty}f_n(x) dx = 0$$
On the other hand,
$$\begin{align}
\sum_{n=1}^{\infty} \int_{\mathbb{R}} f_n(x) &= \sum_{n=1}^{\infty} \int_{\mathbb{R}} (g_n(x) - g_{n-1}(x)) dx \\
&= \sum_{n=1}^{\infty} \left( \int_{\mathbb{R}} g_n(x) dx - \int_{\mathbb{R}} g_{n-1}(x) dx \right) \\
&= \sum_{n=1}^{\infty} (n - (n-1))\\
&= \sum_{n=1}^{\infty} (1) = \infty
\end{align}$$
micromass
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#6
Feb25-13, 09:30 PM
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Of course the Lebesgue integral is much better behaved. In fact, we use the Lebesgue integral especially to have theorems which allow you to switch integral and limit (or series).

There are two important theorems in Lebesgue theory. Those are the monotone and the dominated convergence theorem.

Monotone convergence states that you can switch series and sum if [itex]f_i(x)\geq 0[/itex] for all [itex]x[/itex].
The dominated convergence theorem tells you that you can switch if [itex]\sum_{i=0}^{+\infty} |f_i(x)| \leq g(x)[/itex] for a certain function g such that [itex]\int g<+\infty[/itex]. (We can take g equal to the series of absolute values here)
the_wolfman
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#7
Feb26-13, 12:01 AM
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Opps my bad. I agree that your series should be convergent.


Jbunnii in your example is it true that in the limit of N going to infinity that gn = 0 for all x. Naively I'd expect some sort of singular behaviour at x=0.

If I integrate the sum of f_n from n=1 to N, then I'd get N for all finite N. If I then take the limit as N goes to infinity I'd get infinity not zero.
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Feb26-13, 12:21 AM
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Quote Quote by the_wolfman View Post
Jbunnii in your example is it true that in the limit of N going to infinity that gn = 0 for all x. Naively I'd expect some sort of singular behaviour at x=0.
Let me describe ##g_n(x)## more carefully. Picture an isosceles triangle, whose base is exactly ##[0, 1/n]##, and whose height is ##2n^2##. The coordinates of the three vertices are ##(0, 0)##, ##(1/n, 0)##, and ##(1/(2n), 2n^2)##. I could write a piecewise formula for this but it would just make it more confusing. The function is zero for all ##x \leq 0## and for all ##x \geq 1/n##. The only nonzero portion is in the interval ##(0,1/n)##.

I claim that ##\lim_{n \rightarrow \infty} g_n(x) = 0## for all ##x \in \mathbb{R}##. This is certainly true for all ##x \leq 0## and all ##x \geq 1## because all of the ##g_n## are zero there. All that remains is to check ##(0,1)##. Choose any ##x \in (0,1)##. There exists an ##N## such that ##1/n < x## for all ##n \geq N##. Therefore ##g_n(x) = 0## for all ##n \geq N##, so certainly ##\lim_{n \rightarrow \infty}g_n(x) = 0##. We have shown that this limit is true for all real ##x##.
If I integrate the sum of f_n from n=1 to N, then I'd get N for all finite N. If I then take the limit as N goes to infinity I'd get infinity not zero.
Yes, I don't claim that ##\lim_{n \rightarrow \infty} \int_{\mathbb{R}}g_n(x) dx = 0##. Indeed, this limit is infinite as you said. But the sequence of functions does converge pointwise to zero.
pwsnafu
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Feb26-13, 12:21 AM
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Quote Quote by the_wolfman View Post
Jbunnii in your example is it true that in the limit of N going to infinity that gn = 0 for all x. Naively I'd expect some sort of singular behaviour at x=0.

If I integrate the sum of f_n from n=1 to N, then I'd get N for all finite N. If I then take the limit as N goes to infinity I'd get infinity not zero.
I'm not sure what you are arguing. It's the same as the second half of jbunniii's post.
Mute
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#10
Feb26-13, 02:21 PM
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To continue this topic a little bit, I thought of a related question which the more rigorous mathematicians may be able to answer. Often in physics we encounter cases in which swapping the order of the integral and sum is invalid, but the result is an asymptotic series rather than absolute nonsense. Does anyone know under what conditions swapping integrals and sums gives a legitimate asymptotic series?


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