# I am looking for a disproof of this conjecture on closed forms

Assume $I=\int_0^\infty f(x)\text{ d}x$ and $J=\sum_{n=0}^\infty f(n) ; I,J\in\Bbb{R}$

Conjecture:

If $I$ has a closed form, then $J$ must carry a closed form.

Can someone find a proof or disproof of this conjecture?

I have only managed to prove that this exists for all polynomial functions, but that is fairly elementary. Proving the generalization has remained extremely difficult.

(Note that closed form means Chow's definition of a closed form.)

Edit: The only real exception that I have seen so far is that $f(x)$ can not be in the form $\frac{1}{P(x)}$, where $P(x)$ is a generalized polynomial.

A striking example of this in action is with the function $f(x)=e^{-x^2}$

$I=\sqrt \pi$ and $J=\frac{1}{2}+\frac{1}{2}\theta_3(0,e^{-1})$

with my source being W|A: http://www.wolframalpha.com/input/?i=sum+from+0+to+inf+e%5E-x%5E2

• In principle, changing $f$ on $\Bbb{N}$ will not change the integral, but will change the series. So you need some additional assumptions. – PhoemueX Feb 27 '16 at 3:01
• Side Note 1: Oh, I see (you are looking at substitution). As long as the bounds of both the integral and the series in question after substitiution are still zero and infinity, then the conjecture still applies (I believe). – user311151 Feb 27 '16 at 3:04
• Side Note 2: Note the question: If the integral I does not carry a closed form, that says nothing of the series. – user311151 Feb 27 '16 at 3:05
• What I meant is: Take $f(x)=0$ for all $x\neq 1$ and $f(1)=x_0$, where $x_0$ is a number which does not admit a closed form. Then the integral is $0$ (closed form), but the series is $x_0$ (not closed). – PhoemueX Feb 27 '16 at 3:12
• Ah I see like this question: " math.stackexchange.com/questions/1587356/… " I will have to think about it, but good point there. I feel like that is the only exception to that rule however, so I think that might be the only issue with my conjecture. Thank you very much – user311151 Feb 27 '16 at 3:14