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This question is partially inspired by Qiaochu Yuan's answer to "Will moving differentiation from inside, to outside an integral, change the result?".

Essentially, I would like to know, if we have:

$$\sum_{x=a}^b{ \int{ f(x)dx } } \quad\text{ or }\quad \sum_x{ \int{ f(x)dx } }$$

If we then perform a substitution $\int{ f(x)dx } \to \int{ g(y) dy}$, when can we carry the substitution over to the summation? In other words, when does

$$\sum_{x=a}^b{ \int{ f(x)dx } } = \sum_{y=a_2}^{b_2}{ \int{ g(y)dy } }$$

I'm hoping that someone can cover as much as possible.

I'm wondering how we would derive the new constants of summation as well.

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Matt Groff
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  • Sorry, but I still don't see clearly how the parameters a,b have to see with f(x), g(x) respectively. – user99680 Jun 02 '14 at 00:31
  • @user99680: Ok, so we start with $\int{ f(x) dx}$. Then we get a result, say $h(x)$, from the integral. Now we sum this result, in other words, we take $\sum_{x=a}^b{ h(x) }$. Now, my question is about what happens when we make substitute the integral $\int{ f(x) dx}$ with $\int{ g(y) dy}$. In other words, we make the substitution $x \to j(y)$ for some function $j(y)$. If we do this, we may need to change the limits of substitution, since we are now summing over $y$. So I'm really wondering if this is possible, and what the new $y$ values will be. – Matt Groff Jun 02 '14 at 00:40
  • Sorry to take so much time to get back: is $h(x)$ a discrete function that you are summing? Otherwise, how do you define the sum over a countable set? – user99680 Jun 02 '14 at 05:18
  • @user99680: Yes. We obtain $h(x)$ after the original integration. However, I'm wondering if we substitute in the original integral, how this would affect the summation. Just for your information, it may not be possible to do this. – Matt Groff Jun 02 '14 at 10:40

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The question doesn't really make sense. $x$ is the "dummy variable" in the integration, so $\int_c^d f(x)\; dx$ is not a function of $x$. Thus $$ \sum_{x=a}^b \int_c^d f(x)\; dx = (b-a+1) \int_c^d f(x)\; dx$$

EDIT: For the new version of the problem, let $F(x) = \int f(x)\; dx$ and $S = \sum_{x=a}^b F(x) = F(a) + F(a+1) + \ldots + F(b)$ where $a$ and $b$ are integers, $a < b$. A change of variables is a substitution $x = j(y)$, so that $F(x) = F(j(y)) = G(y)$ where $G = F \circ j$. Thus $F' = f$ and $G'(y)= g(y) = j'(y) f(j(y))$. For $S$ we have $S = G(y_a) + \ldots + G(y_b)$ where $j(y_x) = x$. In general there's no reason for these $y_x$ to be consecutive integers. Of course there are special cases such as $j(x) = x + constant$.

Robert Israel
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