Lucian asked about the almost-integer $2\pi+e\approx9$ in a comment to a partially answered why question about $e\approx H_8$. This is more involved than approximations to $\pi$ and logarithms because two transcendental constants are included, as in $e^\pi-\pi\approx20$.

Tried so far

An answer can be crafted from integrals related to $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$

$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx =\frac{22}{7}-\pi$$

$$\frac{1}{14}\int_0^1 x^2(1-x)^2e^xdx=e-\frac{19}{7}$$

to obtain

$$\int_0^1 x^2 (1-x)^2 \left(\frac{e^x}{14}-\frac{2 x^2 (1-x)^2}{1+x^2}\right) dx = 2\pi+e-9 $$

The visual representation of this integral provided by WolframAlpha shows that $2\pi+e-9$ is positive and small (the integrand is between $0$ and $0.004$ for $0<x<1$), although this is not immediate from the analytic expression.

Moreover, two maxima appear, instead of the single one that is usual in this type of integrals.


Is there a simpler integral with positive integrand in (0,1) that proves $2\pi+e\approx 9$?

  • $\begingroup$ $$\int_0^1 x^4 \left(1 - x\right)^4 \left(\frac{2}{1 + x^2} -\frac{e^x}{24024} \right) dx = 9+\frac{4}{1001} - e - 2\pi$$ has one single maximum but the constant is not exactly 9. wolframalpha.com/input/… $\endgroup$ – Jaume Oliver Lafont Apr 25 '17 at 17:01
  • $\begingroup$ This can be used to write a double inequality $9<2\pi+e<9+\frac{4}{1001}$ and suggests a question for a closer upper bound. $\endgroup$ – Jaume Oliver Lafont May 3 '17 at 15:05

Approximations $e\approx \frac{163}{60} $ and $ \pi \approx \frac{377}{120}$ are related to the integrals

$$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$


$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi.$$

Combining both, we can build $$\frac{1}{2} \int_0^1 (1 - x)^2 \left(e^x - 1 - x - \frac{x^2}{2} - \frac{2 x^5 (1 - x)^4}{1 + x^2}\right) dx = 2\pi+e-9,$$

which explains the result with nonnegative small integrand and a single maximum in $(0,1)$.

WolframAlpha link

This sets the lower bound


For an upper bound, we may take the integral from a failed attempt to match $9$

$$\int_0^1 x^4(1-x)^4\left(-\frac{e^x}{24024}+\frac{2}{1+x^2}\right) dx = 9+\frac{4}{1001}-2\pi-e$$ and write




In short, $2\pi+e$ is close to $9$ because $\pi\approx\dfrac{377}{120}$ from above, $e \approx \dfrac{163}{60}$ from below and $$2·\frac{377}{120}+\frac{163}{60}=9$$

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