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This is a pair of similar questions.

Here is the first one:
Does there exist a rational function $P(x)$ with a,b ∈ ℚ such that $\int_a^b P(x)\operatorname{d}x$ is equal to $k\cdot e$ for some $k\in\Bbb{Q}$ where $e$ is Euler's number? Prove or disprove that such a function exists. If such a function exists find it. Also, this function should be bounded in the y dimension also so there should exist some r such that |P(x)| < r for every x on [a,b].

This is the second:
Let $A$ be a region defined by the inequality $0 < P(x,y)$ where $P$ is a rational function and $x^2+y^2 < r$ for all $(x,y)\in A$ for some finite $r$. Does such an $A$ exist such that the area of $A$ equals $k\cdot e$ where $k\in\Bbb{Q}$ and $e$ is Euler's constant? Prove or disprove that such an $A$ exists. If such an $A$ exist find $P$.

Inspiration for the questions:
A nice way to approximate pi is by picking random points in the region $0 < x < 1$ and $0 < y < 1$ and seeing if $x^2+y^2 < 1$. This is easy to program because all the functions involved are rational. similar things can be done with $\ln(2)$ for example. I would like to do so with $e$ but can't think of the correct shape to use.

The coefficients of the rational functions should be rational numbers. The bounds on a and b in the first question should be rational.

I have now restated this question more clearly Creating e using rational functions. Thanks for the help fixing up the wording.

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  • $\begingroup$ You might want to ask for the coefficients of the nominator and denominator of $P$ to be rational numbers... $\endgroup$ – user378947 Dec 12 '16 at 4:13
  • $\begingroup$ I think the question you wish to ask has more restrictions on it than that: choose $P(x)=1, a=0, b=e$, for instance, or $P(x,y)=\frac{e}{\pi}, r=1$. $\endgroup$ – RavenclawPrefect Dec 12 '16 at 4:13
  • $\begingroup$ I'm not interesting in Pi for this question I was just mentioning how I came up with the problem. I'm confused as to what a and b are in your comment. $\endgroup$ – mathew Dec 12 '16 at 4:23
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    $\begingroup$ Do you want $a,b$ and the coefficients to be rational? $\endgroup$ – John Wayland Bales Dec 12 '16 at 4:50
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    $\begingroup$ OK. The question doesn't indicate $a$ and $b$ must be rational. $\endgroup$ – alex.jordan Dec 12 '16 at 4:58
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I'm assuming $a$ and $b$ are fixed rational numbers.

If $P(x) = N(x)/D(x)$ is a rational function with rational coefficients for numerator and denominator, then the roots of the denominator $D(x)$ are algebraic numbers. We can perform a partial fraction decomposition of $P(x)$ and integrate it from $x=a$ to $x=b$, obtaining a result that is a linear combination of $1$ and natural logs of algebraic numbers with algebraic coefficients. The question is whether $e$ can be expressed in this way.

Unfortunately this is still an open question, though I believe it would follow from Schanuel's conjecture that it can't be so expressed. We don't even know that $e + \pi$ is irrational. If $e + \pi = r$ was rational, you could write $$e = \int_0^1 \left(r - \frac{4}{1+x^2}\right)\; dx$$

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  • $\begingroup$ Thanks for the interesting answer but I'm looking for something I could implement so without knowing r I can't actually use this. Very nice trick though. $\endgroup$ – mathew Dec 12 '16 at 5:10
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    $\begingroup$ I think you misunderstand: the point is that almost certainly you can't do this, although we can't prove that you can't. $\endgroup$ – Robert Israel Dec 12 '16 at 5:32
  • $\begingroup$ ooh, didn't realize that, so are you saying that if e+pi is irrational that this is impossible? $\endgroup$ – mathew Dec 12 '16 at 5:40
  • $\begingroup$ If $e + \pi$ is irrational then that particular rational function doesn't work. If Schanuel's conjecture is true it is impossible. $\endgroup$ – Robert Israel Dec 12 '16 at 7:24

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