# “closed form” finite sum

If a finite sum has a definite integral representation, for which it can be proved the underlying indefinite integral is not an elementary function, then does this imply the original finite sum can not be expressed as an elementary function (on applying the bounds of the original representation). Or does this mean that the method of integration is a dead end in this case?

I suppose another away of asking this question is: Do there exist finite sums which can be expressed as an elementary function (without summation signs), but for which there also exist integral representations whose associated indefinite integral is non-elementary?

-
So, for example, if your finite sum can be shown to equal $$\int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2},$$ then you want to conclude that the original sum is not expressed as an elementary function? –  Gerald Edgar Mar 30 '12 at 14:30
What??? $\sum_{i=1}^1 \frac{\pi}{2} = \pi/2.$ –  Igor Rivin Mar 30 '12 at 15:10
So: edit the question to be what you actually want! As Igor showed, any number is a "finite sum" with one term. –  Gerald Edgar Mar 30 '12 at 15:35
My interpretation is that the question is asking whether there is an elementary function with an elementary indefinite summation but not indefinite integral. I.e., if $f(x)$ is elementary and $g(x+1)-g(x) = f(x)$ has an elementary solution $g(x)$, then must $h'(x)=f(x)$ have one as well? I'm sure the answer is no. For example, if $g(x) = \exp(-x^2)$, then this amounts to asking whether $\textup{erf}(x+1)-\textup{erf}(x)$ is an elementary function. Presumably it isn't, but I don't know a proof offhand. I imagine enough about differential algebra is known to prove this, though... –  Henry Cohn Mar 30 '12 at 23:09
$g(x) = \exp(i \pi x)/x$ is an elementary solution to $g(x+1) - g(x) = - \exp(i \pi x) \frac{2x+1}{x^2+x}$, and it is straightforward to use Liouville's criterion: $u e^v$ (where $u$ and $v$ are rational functions of $x$) has an elementary antiderivative iff the differential equation $y' + v' y = u$ has a rational solution. But in this case $u$ has a pole of order $1$ at $x$, and if a rational solution $y$ had a pole of order $k \ge 1$ there, $y' + 2 i y$ would have a pole of order $k+1$. –  Robert Israel Mar 31 '12 at 1:05