$\def\d{\mathrm{d}}$If there is an explanation for $e^\pi-\pi \approx 20$ similar to this one for $2\pi+e \approx 9$ we may try to build it the same way, from integrals related to rational approximations to $e^\pi$ and $\pi$.

Several integrals link $\pi$ to close fractions, such as Dalzell-type integrals for convergents to $2\pi$, but how about $e^\pi$?

A failed attempt

One way to describe $e^\pi$ as a fraction plus an integral error would be given by the following integral basis:

$$ \int_0^1 \frac{4e^{4\arctan x}}{1+x^2} \,\d x=e^\pi-1,\\ \int_0^1 \frac{\left(1+4x+x^2\right)e^{4\arctan x}}{1+x^2}\,\d x=e^\pi. $$

However, this leads to integrands that change their sign in $(0,1)$ and are not small, so they are not useful as a proof.

For instance, for $e^\pi \approx 23$, that we could combine with $\pi\approx 3$ to reach $e^\pi-\pi \approx 20$, we have $$\int_0^1 \frac{\left(70-88x-22x^2\right)e^{4\arctan x}}{1+x^2}\,\d x=e^\pi-20.$$

WolframAlpha link

Similarly, for $e^\pi\approx \dfrac{162}{7}$ there is

$$\int_0^1\frac{(155 x^2 + 620 x - 493)e^{4\arctan x}}{7(1+x^2)} \,\d x = \frac{162}{7}-e^\pi.$$

WolframAlpha link

The graphs by WolframAlpha show sign changes in $(0,1)$. Therefore, different integrals should be found.


Are there integrals with small nonnegative integrand that yield rational approximations to $e^\pi$?

Also related:

Rational series representation of $e^\pi$

Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$?


1 Answer 1



$\displaystyle J(a,b,c)=\int_0^1 \dfrac{(ax^2+bx+c)\text{e}^{4\arctan (x)}}{1+x^2}dx$

you want to find $a,b,c>0$ integers and $\alpha,\beta,\gamma$ integers such that:

$\alpha J(a,b,c)+\beta \text{e}^{\pi}+\gamma=0$

Function lindep from GP PARI can help to find empirically such integers.

you can use a triple loop for $0<a,b,c<max$. You compute J(a,b,c) and then you invoke lindep to find $\alpha(a,b,c),\beta(a,b,c),\gamma(a,b,c)$. If these numbers are not too big, you win.

PS: Condition on a,b,c need to be fixed. $a>0$ and $b^2-4ac<0$


It seems that $\boxed{J(1,4,5)-2\text{e}^{\pi}+1=0}$ (related to $x^2+4x+5$)

An antiderivative for $\dfrac{(x^2+4x+5)\text{e}^{4\arctan x}}{1+x^2}$ is,

$(1+x)\text{e}^{4\arctan x}$

(actually there are tons of J(a,b,c) that seem working. GP PARI has found plenty of them in few seconds)

  • 1
    $\begingroup$ if $\Delta<0$ and $a>0$ it seems that $J(a,b,c)\geq \dfrac{\text{e}^\pi-1}{4}\cdot\dfrac{-\Delta}{4a}$ $\endgroup$
    – FDP
    Apr 26, 2017 at 22:06
  • $\begingroup$ Thank you! The integrand is nonnegative, but the approximation $e^\pi\approx \frac{1}{2}$ does not seem to make much sense, because $e^\pi > 23$. The integrand should be smaller, say $<1$ in $(0,1)$. For this function, it is larger than 100 at x=1 (wolframalpha.com/input/…) I don't claim that changing $a,b,c$ in the integral written is enough, possibly a different structure is needed... $\endgroup$ Apr 27, 2017 at 11:03
  • $\begingroup$ Yep i know. As i told you there are plenty of such J(a,b,c) (but integrals are not small). It will be hard to put together all these conditions: small a,b,c, $ax^2+bx+c>0$ and J(a,b,c) small. (Read my other comment) $\endgroup$
    – FDP
    Apr 27, 2017 at 13:53
  • $\begingroup$ I think small $a,b,c$ is not required, is it? $\endgroup$ Apr 27, 2017 at 17:05
  • 2
    $\begingroup$ You're right but it was easier to program $\Delta<0$ and $a>0$ aimed to search for $J(a,b,c)$. $0<a,b,c\leq 100$, $\Delta<0$, $J(a,b,c)<5$ gave me nothing. $\endgroup$
    – FDP
    Apr 27, 2017 at 21:38

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