Is there a definite integral for which the Riemann sum can be calculated but for which there is no closed-form antiderivative?

Some definite integrals, such as $\int_0^\infty e^{-x^2}\,dx$, are known despite the fact that there is no closed-form antiderivative. However, the method I know of calculating this particular integral (square it, and integrate over the first quadrant in polar coordinates) is not dependent on the Riemann sum definition. What I thought might be interesting is a definite integral $\int_a^bf(x)\,dx$ for which the limit of the Riemann sums happens to be calculable, but for which no closed-form antiderivative of $f$ exists. Of course there are some obvious uninteresting examples, like integrating odd functions over symmetric intervals, but one doesn't need Riemann sums to calculate these uninteresting examples.

Edit: To make this a bit clearer, it would be nice to have a "natural" continuous function $f(x)$ where by some miracle $\lim_{n\to\infty} \sum_{i=1}^nf(x_i)\Delta x$ is computable (for some interval $[a,b]$) using series trickery, but for which no antiderivative exists composed of elementary functions.

This may not answer your question, but I read of another example of an integral which can be calculated without finding an anti-derivative in Halmos's Problems for Mathematicians, Young and Old is the integral: $$ \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dx dy.$$ The trick here is to look at the integrand at each $x,y$ as a number $\frac{1}{1-r}$ where $0<r=xy<1$ and follow your heart.

Edit:

On second thought, this answer is probably not to the point, since the solution will require a convergence theorem (monotone convergence works), and then we need an anti-derivative-less calculation of integrals of the form $\int_{0}^{1} x^k dx$. There are such (found here, for instance). In any case, the method for finding this integral invites the generalizations$$\int_{0}^{1} \cdots \int_{0}^{1} \int_{0}^{1} \frac{1}{1- x_1 x_2 \cdots x_n} dx_1 dx_2 \cdots dx_n$$ where $n>1$.

Dirichlet function defined on $\displaystyle [0,1]$:

$\displaystyle D(x) = 0$ is x is irrational.

$\displaystyle D(x) = \frac{1}{q}$ if $\displaystyle x = \frac{p}{q}$ in a rational in reduced form, $\displaystyle q \gt 0$.

$\displaystyle F(x) = \int_{0}^{x} \ D(t) \text{d}t = 0 \ \forall \ x \in [0,1]$

Note that there is no function $\displaystyle f$ such that $\displaystyle f'(x) = D(x)$, because derivatives have the intermediate value property (see here for instance: http://www.math.wvu.edu/~kcies/teach/Fall03Spr04/451NadlerText/451Nadler101-120.pdf) and $\displaystyle D(x)$ takes only rational values.

Not really sure if this is what you are looking for, though.

If one can prove that an integral is some value, you have proven that the limit of Riemann Sums is the same value.

Anyway, hope that helps.