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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.

Cheerful Parsnip
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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? – Qiaochu Yuan Feb 25 '11 at 19:30
  • 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.) – Cheerful Parsnip Feb 25 '11 at 20:01
  • 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. – Qiaochu Yuan Feb 26 '11 at 01:33
  • @Qia: And I was guessing the answer will be "yes" :-) – Aryabhata Feb 26 '11 at 01:39
  • 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? :-) – ShreevatsaR Feb 26 '11 at 05:25
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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 $e^{-x^2}$. – Jonas Meyer Feb 26 '11 at 05:58
  • Jonas is correct. – Cheerful Parsnip Feb 26 '11 at 12:25

2 Answers2

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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 $n>1$.

AppliedSide
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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.

Aryabhata
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  • I agree that this is a (classic) example, but it's not really what I was looking for. I was hoping for a ``natural" function $f$ for which an antiderivative exists, but is just not useful computationally. (An antiderivative for $e^{-x^2}$ exists but not using standard functions.) – Cheerful Parsnip Feb 25 '11 at 19:22
  • @JIm: Yeah I guessed. Perhaps one can convert the proofs of some of the well known integrals into statements in terms of Riemann Sums... – Aryabhata Feb 26 '11 at 01:31