21
$\begingroup$

In $$\sum_{x \in [0,1]} e^x,$$ $e^x$ is summed over all values in the interval $[0,1]$.

Am I right to say that $$\sum_{x \in [0,1]} e^x = \int^{x=1}_{x=0} e^x \, \mathrm dx?$$

$\endgroup$
27
$\begingroup$

No, they're not equal; the sum of uncountably many positive numbers is necessarily infinite, whereas the integral $$\int_0^1\exp(x)\,dx=e-1$$ is finite.

$\endgroup$
30
$\begingroup$

Since nobody has wrote about this yet, let me mention that this sum indeed represents an integral, but not the integral you thought of. Namely, for any set $X$ and any function $f:X\to\Bbb R$ $$ \sum_{x\in X}f(x) = \int_Xf(x)\mu(\mathrm dx) $$ where $\mu$ is the counting measure. In your case $X = [0,1]$, and you instead thought of integrating with a more natural Lebesgue measure $\lambda$ which is of course different from the counting one, so $$ \sum_{x\in[0,1]} f(x) = \int_{[0,1]}f(x)\mu(\mathrm dx)\neq \int_{[0,1]}f(x)\lambda(\mathrm dx) $$ in general. In particular, the inequality does not hold for $f(x) = \exp (x)$ as other answers showed.

In fact, both integrals only agree (being finite) if they both are zero. Indeed, if the counting integral is non-zero and finite, then $f$ takes only countably many non-zero values - but then the Lebesgue integral is zero. Vice-versa, if the Lebesgue integral is non-zero, then $f$ takes uncountably many values so that either the counting integral is undefined, or infinite.

$\endgroup$
  • $\begingroup$ @Legendre: thanks. I've added some details on when both integral equal each other. $\endgroup$ – Ilya Jul 11 '13 at 15:04
  • 1
    $\begingroup$ This is very nice. $\endgroup$ – Alexander Gruber Jul 12 '13 at 18:18
7
$\begingroup$

Not exactly. Taken literally, $\sum_{x\in[0,1]}\exp(x)$ is a sum with uncountably many positive summands and cannot evaluate to a finite value. (For example, thee are more than $10^{100}$ summnds and all are $\ge1$, hence the sum is at least $>10^{100}$).

$\endgroup$
  • $\begingroup$ yep, but $10^{100}$ is finite. $\endgroup$ – ThomasMcLeod Jul 11 '13 at 19:36
  • 3
    $\begingroup$ @ThomasMcLeod Sure, but this holds for every $10^{100}$, as large as one wants. $\endgroup$ – Did Jul 13 '13 at 11:34
6
$\begingroup$

Do these Inequalities give an idea ?

$$\sum_{x\in[0,1]}\exp(x)\geq \sum_{x\in[0,1]}1$$ and $$\int_0^1 \exp(x)dx\leq e\int_0^1 dx$$

$\endgroup$
1
$\begingroup$

$\sum_{x\in[0,1]}\exp(x)>\sum_{x\in[0,1]}x>\sum_{n\in\mathbb{N}}\frac{1}{n}$

Since $\sum_{n\in\mathbb{N}}\frac{1}{n}$ diverges so does $\sum_{x\in[0,1]}\exp(x)$

$\endgroup$
0
$\begingroup$

You can't sum "over all values", but what you can do is consider a partition of $[0,1]$ into infinitely many subintervals where each subinterval has infinitesimal length, which can be denoted $dx$. More precisely, one chooses an infinite integer $N$, and partitions the interval into $N$ equal parts, by means of partition points $x_i$. Then one takes the infinite sum $\sum_{i=0}^{N-1} e^{x_i} dx$. To obtain the integral, one then takes the standard part.

$\endgroup$
  • $\begingroup$ an "infinite integer" ?? $\endgroup$ – Zev Chonoles Jul 11 '13 at 14:47
  • $\begingroup$ See en.wikipedia.org/wiki/Hyperinteger and enjoy! $\endgroup$ – Mikhail Katz Jul 11 '13 at 14:49
  • $\begingroup$ Okay, I'll accept that it's possible to define integration this way using nonstandard analysis (though I think the fact that this is not the usual construction of the integral should be mentioned), but why do you say that one can't sum over all values? $\endgroup$ – Zev Chonoles Jul 11 '13 at 14:52
  • $\begingroup$ Note that one can indeed have an uncountable sum of hyperreals that's not infinite, such as the sum above (of course with reals this does not work). I think the fact that Leibniz, Euler, and Cauchy constructed the integral along these lines should be mentioned. As far as the sum "over all values", it can formally be imagined but does not lead to anything useful; in this sense one "can't sum over all values". $\endgroup$ – Mikhail Katz Jul 11 '13 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.