Can $\sum_{x \in [0,1]} e^x$ be represented as an integral?

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 $$\sum_{x \in [0,1]} e^x = \int^{x=1}_{x=0} e^x \, \mathrm dx?$$

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $>10^{100}$).

Solution 4:

Do these Inequalities give an idea ?

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

Solution 5:

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

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