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 $$2·\frac{377}{120}+\frac{163}{60}=9$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.