
#1
Feb2513, 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! 



#2
Feb2513, 07:40 PM

P: 135

Always.
∫ a+ b dx= ∫ a dx+ ∫ b dx 



#3
Feb2513, 08:01 PM

HW Helper
P: 1,391





#4
Feb2513, 09:03 PM

Sci Advisor
P: 779

integrating infinite sums 



#5
Feb2513, 09:18 PM

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PF Gold
P: 2,927

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_{n1}(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, $$\begin{align} \sum_{n=1}^{\infty} \int_{\mathbb{R}} f_n(x) &= \sum_{n=1}^{\infty} \int_{\mathbb{R}} (g_n(x)  g_{n1}(x)) dx \\ &= \sum_{n=1}^{\infty} \left( \int_{\mathbb{R}} g_n(x) dx  \int_{\mathbb{R}} g_{n1}(x) dx \right) \\ &= \sum_{n=1}^{\infty} (n  (n1))\\ &= \sum_{n=1}^{\infty} (1) = \infty \end{align}$$ 



#6
Feb2513, 09:30 PM

Mentor
P: 16,692

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) 



#7
Feb2613, 12:01 AM

P: 135

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. 



#8
Feb2613, 12:21 AM

Sci Advisor
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PF Gold
P: 2,927

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##. 



#9
Feb2613, 12:21 AM

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#10
Feb2613, 02:21 PM

HW Helper
P: 1,391

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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