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So, in general, I can get this value:

$$\lim_{a \to \text{a constant}}{ \int{ \left( \sum_{x=x_1}^{x_2}{ f(a,x) }da \right)} } \tag{1}$$

What I'm after is this:

$$\sum_{x=x_1}^{x_2}{\left( \lim_{a \to \text{a constant}}{ \int{ f(a,x) }da } \right)} \tag{2}$$

So, essentially, I want to move the sum from inside an integral to outside an integral. The integral is very, very simple - it's continuous and Riemann integrable, etc. It can be done with elementary calculus. Then I plug in a value for the integrand.

The sum is also fairly simple; it is finite. The problem is that if I integrate first, I can't solve the resulting sum as it is. So I want to exchange the sum and the integral so that I can effectively get this value.

WHAT I'M AFTER

I'd like to know if I can always exchange these. Or, if not, when I can exchange these. I've already plugged in some values into my math software and both (1) and (2) gave the same result. I want to prove when I can do this, so I really also need sources or references that explain when I can do this.

SOME REFERENCES

Exchanging Sums and Integrals: Can we possibly exchange summation and integration with negative values?

Exchanging Sums and Limits: When can we exchange limits and summations?

I REALLY NEED A FAIRLY THOROUGH EXPLANATION BECAUSE I WANT TO PROVE THIS.

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Matt Groff
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    The exchange is valid provided that you have a finite sum. For infinite sums, you would need to justify it. – Joel Sep 24 '14 at 15:08
  • @Joel: Just as I thought. I want to prove this - in other words, I want to be fairly rigorous. However, I'm guessing that this is just a basic property of finite sums. Is there any way I can prove this fairly simply? Again, maybe this is just a property of a finite sum, since there's no need to worry about how the terms can be rearranged... Thinking about it some more, I guess this is justified because $\int{(a + b)} = \int{a} + \int{b}$. – Matt Groff Sep 24 '14 at 15:14
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    You are correct Matt. It follows from the linearity of limits and integrals. Linearity can come into play for finite sums. It does not always work for infinite sums since that is a limit of finite sums, so the interchange of limits gets in your way. – Joel Sep 24 '14 at 16:08

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