# Why do people interchange between $\int$ and $\sum$ so easily?

One of the things I found curious in many texts is how in certain cases interchange the $\sum$ operator with $\int$. What are the "terms" for such a swap? I understand that integration in the early days was seen as an approximation of the area under the curve by using the very definition of multiplication and area to lend a hand with very small increments where the number of samples goes to infinity.

Beyond the original question, is this also the reason why we keep the right hand $dx$ (or any other infinitesimal variable), just to remind us of the origin because it "multiplies against the function", hence giving area. Or is there more to it?

Hints, answers, references to books... I'd appreciate anything you can give me.

• Do you mean $\sum \int f_i\, dx = \int \sum f_i\, dx$ as in here or $\int f_i\, dx = \sum f_i \Delta\,x$ as in here?
– user2468
Commented Apr 25, 2012 at 15:53
• @J.D. Not the first one (sum rule, right?)... That just an addition for loop, that's justified. It's the second one. I would like to understand that more. Commented Apr 25, 2012 at 16:02
• have a look at en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula which essentially says $\sum = \int + \mathrm{error term}$, where the error term is hopefully controllable for nice functions. A derivation is sketched here math.stackexchange.com/questions/123633/… Commented Apr 25, 2012 at 17:18
• Summation of positive terms is a special case of integral for counting measure. Commented Jun 23, 2012 at 13:11

In elementary analysis, a Riemann/Darboux integral is defined (among other equivalent definitions) as a suitable limit of a (finite) sum. Whence the folklore according to which "an integral is essentially a series". This is rather false, but you know, in elementary analysis/calculus you can almost say whatever you wish.

The $\mathrm{d}x$ is clearly a deformation of $\Delta x$ in Riemann sums. Nowadays, it denotes the measure for which the integral is defined. If the integral is just a Riemann integral, some authors suggest to write $\int_a^bf$ instead of $\int_a^bf(x)\, \mathrm{d}x$. They are right, since the Riemann integral depend on $a$, $b$, and the function $f$. The variable of integration is a dummy one.

Finally, remember that $\int$ is a calligraphic deformation of an "S", while $\sum$ is the greek "S". Hence many pioneers used to kind of confuse $\sum$ and $\int$ in their manuscripts. But, honestly, contemporary textbooks should not swap the two signs, since we live in 2012 and Cauchy died many years ago ;-)

• Very nice, thank you. On another note, your text could benefit from some formatting (a paragraph or two). Commented Apr 25, 2012 at 16:30

IMHO, the integral which is the closest as a notion to the sum is Lebesgue integral. First, a sum is a Lebesgue integral with respect to an appropriate measure, i.e. $$\sum\limits_{i=1}^n a_i = \int\limits_1^n a(x)\;\mu(\mathrm dx)$$ where $a(x)$ is any function with the only restriction $a(i) = a_i$ for $i=1,\dots,n$ and measure $\mu$ is concetrated at points $1,2,\dots,n$ such that $\mu(1) = \mu(2) = \dots = \mu(n)$.

Since the sum is an object with many nice properties, it is always useful when the integral also shows similar properties. E.g. if $a_i\geq 0$ and $$\sum\limits_i a_i = 0$$ then $a_i = 0$ for all $i$. For Lebesgue integral you have almost the same, namely if the function $f$ is such that $f(x)\geq 0$ and $$\int\limits_X f(x)\mu(\mathrm dx) = 0$$ then the set $\{x:f(x)\neq 0\}$ is of zero measure $\mu$.

• But probably the Riemann-Stieltjes is even closer to a sum, as Rudin points out in his celebrated book Principles of mathematical analysis. Commented Jul 3, 2012 at 14:31
• The key point is that interchanging a series and an integral is the same as interchanging two integrals, and Fubini's theorem applies. Commented Jul 3, 2012 at 15:02

I would like to show an example how we can change ∫ to ∑

If a real-valued function $f(t)$ is infinitely differentiable at $0 \leq t \leq x$ and the whole high order derivative values are defined at 0≤t≤x.

Firstly , we can write Maclaurin series of $f(t)$ at point $t=0$ If a real-valued function f is infinitely differentiable at $t=0$ and the whole high order derivative values are defined.

Maclaurin series of $f(t)$ at point $t=0$:

$f(t) =f(0)+\frac{f'(0)t}{1!}+\frac{f''(0)t^2}{2!}+.....=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t ^n$

$$\int _0^x {f(t) dt}=\int _0^x(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n)dt=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\int _0^x t^n dt)=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\frac{x^{n+1}}{n+1})$$ $$\int _0^x {f(t) dt}=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^{n+1}}{(n+1)!}$$ $$(1)$$

$$f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m$$ $$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants. More information about summation http://en.wikipedia.org/wiki/Summation

$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n \sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m=\lim_{n\to\infty} \frac{x}{n}\sum_{m=0}^{\infty} \frac{x^m}{n^m} \frac{f^{(m)}(0)}{m!} \sum \limits_{k=1}^n k^m=\lim_{n\to\infty} \frac{x}{n}[f(0)n+\frac{f'(0)x}{n 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^2}{n^2 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^3}{n^3 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^4}{n^4 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= \lim_{n\to\infty} [f(0)x+\frac{f'(0)x^2}{n^2 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^3}{n^3 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^4}{n^4 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^5}{n^5 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= [f(0)x+\frac{f'(0)x^2}{ 2!}+ \frac{f''(0)x^3}{ 3!}+\frac{f'''(0)x^4}{ 4!}+\frac{f^{(4)}(0)x^5}{ 5!}+...... ]$$

$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)x^{m+1}}{(m+1)!}$$ $$(2)$$

Equation $(1)$ and equation $(2)$ are equal to each other. The proof is completed.

$$\int _0^x {f(t) dt}=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$$

• I have a doubt. For a continuous function $f$, is this not simply the definition of the integral by means of (uniform) Riemann sums? You are simply using partitions whose nodes are $0+\frac{x-0}{n}k$, as $k=0,\ldots,n$. Commented Jul 3, 2012 at 8:19
• More generally: $$\int _a^x {f(t) dt}=\lim_{n\to\infty} \frac{(x-a)}{n}\sum \limits_{k=1}^n f(a+\frac{k(x-a)}{n})$$ en.wikipedia.org/wiki/Fundamental_theorem_of_calculus Commented Jul 3, 2012 at 8:26
• So, your answer is a definition with useless assumptions? ;-) Commented Jul 3, 2012 at 8:29
• It is an example how to change an integral to sum. Why do you think it is useless? Please tell me which point is wrong in my answer. Commented Jul 3, 2012 at 8:32
• No, there is (probably) nothing wrong. But your answer sounds like "Any integral is (the limit of) a sum". The only useless thing is the high regularity of $f$: the formula is actually true whenever $f$ in Riemann-integrable. Ah, one more thing: by definition, $f$ is a Taylor series if and only if it is real analytic. Differentiability at a single point, as you write, is not enough. Commented Jul 3, 2012 at 8:35