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Integrals are defined in terms of series, so why is the treatment of definite integrals different than the treatment of series when the lower limit is greater than the upper limit.

For a definite integral whose lower limit $a > b$ upper limit, we generally accept that

$$\int_a^b f(x)\,dx := - \int_b^a f(x)\,dx \qquad (1)$$,

whereas for a series/sum different whose lower limit $a > b$ upper limit, we generally consider it an empty sum equal to $0$

$$\sum_{i=a}^b x_i := 0 \qquad (2)$$


Since integrals are defined in terms of series (AFAIK), where does this difference arise? Is it just a matter of convention? What's useful?


Empty sum question

Definite integral question


My best guess is

Because we define definite integrals in terms of their antiderivative

$$\int_a^b f(x)\,dx = F(b) - F(a)$$

it is necessary to have the property

$$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$

to satisfy certain desirable properties.

In contrast, series are not defined in terms of some antisummand, so it is not necessary to have an analogous property.

However, from the point of view that integrals are defined in terms of series, this attempt is not fully satisfactory.

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    $\begingroup$ Consider reading the first couple pages of Tao's "Differential forms and integration." A fruitful perspective might be that the summation is over an index set $\{1,\dots,n\}$, which generalizes to the (unoriented!) integral over a set $\int_{[a,b]}$. On the other hand, as soon as we introduce orientatation of $[a,b]$, which is foreshadowed by the notational choice $\int_a^b = -\int_b^a$, the integral departs qualitatively in a significant way from the notion of summation. If we cared about orienting sums (I do not know what this would look like), the analogy would be symmetric. $\endgroup$
    – While I Am
    Jan 29, 2022 at 6:52

2 Answers 2

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I disagree with certain remarks and premises of your question, namely

  1. "Because we define definite integrals in terms of their antiderivative..."

We don't define integrals in terms of antiderivatives. As per your initial characterization, we define them in terms of sums (Riemann, Darboux, etc.). However, the fundamental theorem of calculus links the two concepts together.

  1. "...for a series/sum different whose lower limit $a > b$ upper limit, we generally consider it an empty sum equal to $0$"

That is not the "general" convention (though it may be adopted in some particular situations). A more convincing and consistent convention that is analogous to that adopted for integrals is explained below. Moreover, as you will see, the fact that a sum of the form $\sum_{k=c+1}^{c}$ is assigned a value of zero (as per the question in the link you share, where $c=-1$) is a special case of this more consistent convention, and is directly analogous to integrals of the form $\int_c^c$ being assigned a value of zero.


Recall the property that for $a< b< c$,

$$\int_a^c f(x)dx=\int_a^b f(x)dx+\int_b^c f(x)dx \quad (1)\\ \implies \int_a^b f(x)dx=\int_a^c f(x)dx-\int_b^c f(x)dx.\quad (2)$$

Note that

$$\int_c^b f(x)dx:=-\int_b^c f(x)dx\quad (3)$$

is a convention. This convention is useful since $(3)$ allows $(2)$ to be written as

$$\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx,$$

so the property $(1)$ is preserved regardless of the ordering of $a,b,c.$ Of course, the convention is also intuitive in terms of interpreting the integral as representing signed area.


It's not hard to see the analogous convention can be adopted for sums.

We know for integers $a< b< c,$

$$\sum_{k=a}^c x_k=\sum_{k=a}^b x_k+\sum_{k=b+1}^c x_k\quad (1')\\ \implies \sum_{k=a}^b x_k=\sum_{k=a}^c x_k-\sum_{k=b+1}^c x_k.\quad (2')$$

Adopting the convention

$$\sum_{k=c+1}^b x_k:=-\sum_{k=b+1}^c x_k\quad (3')$$

allows $(2')$ to be written as

$$\sum_{k=a}^b x_k=\sum_{k=a}^c x_k+\sum_{k=c+1}^b x_k,$$

thus preserving $(1')$ regardless of the ordering of $a,b,c.$

You can see that applying the convention $(3')$ to the case $b=c$ implies any sum of the form $\sum_{k={c+1}}^c$ is assigned the value zero. But this convention does not assign zero to all sums with an "upper bound" that is less than the "lower bound." The case where the upper bound is exactly one less than the lower bound is assigned zero for consistency.

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  • $\begingroup$ Thanks. That makes sense. I found the following argument more straightforward: Start with $(2')$ $$\sum_{k=a}^b = \sum_{k=a}^c - \sum_{k=b+1}^c \qquad(2')$$ Let $a = c < b$. Then we have $$ \begin{aligned} \sum_{k=a}^b x_k &= \sum_{k=a}^a x_k - \sum_{k=b+1}^a x_k\\ x_a + \sum_{a+1}^b x_k &= x_a - \sum_{k=b+1}^a x_k\\ \sum_{a+1}^b x_k &= - \sum_{k=b+1}^a x_k &&(3') \end{aligned}$$ $\endgroup$
    – joseville
    Jan 29, 2022 at 18:49
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$\int_a^b f(x)\,dx$ is defined as the signed area under the curve $f(x)$ from a to b. It is the fundamental theorem that equates its value to F(b) - F(a). Clearly, the sign changes when you interchange the limits.

The sum of a series from a to b is by no means equivalent to $\int_a^b f(x)\,dx$. $\sum_{i=a}^b x_i dx$ as $dx$ tends to $0$ is equivalent to $\int_a^b f(x)\,dx$.

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  • $\begingroup$ Shouldn't it be more like $\sum_{i=a}^b f(x_i)dx$? But also, how are $x_i$ and $dx$ defined in this case? $\endgroup$
    – joseville
    Jan 29, 2022 at 18:22
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    $\begingroup$ Here I have taken $x_i$ as the value of the function at the $i^{th}$ position. I took it that way because you were comparing $\sum_{i=a}^b x_i$ with the definite integral. Read about Reimann integral for more clarity Reimann integral $\endgroup$
    – Sasikuttan
    Jan 29, 2022 at 18:59

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