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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.

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    I don't really understand the question. What do you mean when you say that the Riemann sum is computable? Do you mean in the formal sense of computability?2011-02-25
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    No, I don't mean the formal sense of computable. I mean that you can actually figure out the limit. In Calculus we show our students how to find $\int_0^1x^2\,dx$ by actually taking a limit of a Riemann sum using the identity $1^2+\cdots+n^2=(1/6)n(n+1)(2n+1)$, but later it turns out we could get the answer by finding an antiderivative ($x^3/3$.) My question is whether the former step can be solved for some $f(x)$ but not the latter step. (As in writing down an actual answer.)2011-02-25
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    Ah. I think I understand, but I would still appreciate a definition formal enough that it would be possible to answer "no" to this question.2011-02-26
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    @Qia: And I was guessing the answer will be "yes" :-)2011-02-26
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    If for the Riemann sums you get some closed-form expression (in terms of Δx, a, and b), then taking the limit of that expression as Δx→0 will give the definite integral in closed-form. So if you have a closed form for the definite integral (in terms of a, b), then it seems there's a closed-form antiderivative as well (put a=0). So the answer should be No, unless you pick a function and interval such that the Riemann sums exist *but* either there's no limit for the expression as Δx→0 or you simply cannot put a=0 for some reason. Can such functions be "natural" enough for you? :-)2011-02-26
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    @ShreevatsaR: I took it to mean that the limit of Riemann sums has a nice closed form expression for a particular interval, not in any way that indicates how it varies as the endpoints of the interval change. I.e., the closed form expression would just be a number, just like we know what $\int_0^\infty e^{-x^2}dx$ is without having knowledge of a "closed form" for an antiderivative of $e^{-x^2}$.2011-02-26
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    Jonas is correct.2011-02-26

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