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Recently I noticed this integral:

$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}7-\pi\approx0$$

Which is a very interesting result which gives us the value of $\pi\approx\frac{22}7\approx3.142857142$ with 2 decimal places correct.

Note that actual value of $\pi\approx$3.14159265359

  • I suppose this happens because as $0<x<1$. So $\displaystyle \frac{x^4(1-x)^4}{1+x^2}\ll1$, so integral must approximately be zero.
  • Integrating this by hand: $$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\int_0^1\left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{x^2+1}\right)dx$$ which can be easily calculated, and the $(x^2+1)^{-1}$ term will make an $\arctan$ term which will generate $\pi$.
  • Taking this to the next level I calculated these which all can be done by hand:

$$\int_0^1\frac{x^8(1-x)^8}{1+x^2}dx=4\pi-\frac{188684}{150115}\approx0\quad:\pi\approx\frac{188684}{4\times150115}=\frac{47171}{15015}\approx3.141591741$$

  • 5 decimal places correct.

$$\int_0^1\frac{x^{12}(1-x)^{12}}{1+x^2}dx=\frac{431302721}{8580495}-16\pi\approx0\quad:\pi\approx\frac{431302721}{16\times8580495}=\frac{431302721}{137287920}\approx3.14159265433$$

  • 8 decimal places correct.

  • With such an effort(probably using some software) we can go to these:

$$\int_0^1\frac{x^{8n}(1-x)^{8n}}{1+x^2}dx=A(n)-B(n)\pi\tag{1}$$ $$\int_0^1\frac{x^{8n+4}(1-x)^{8n+4}}{1+x^2}dx=C(n)\pi-D(n)\tag{2}$$ where $n={0,1,2,3,\ldots}$ and $A,B,C,D$ are functions of n, all of which are always positive.

  • And with $n\to\infty$ we'll probably reach exact value of $\pi$

Results:

  • This type of integrals $(1)$ and $(2)$ are very close to zero and help finding values of $\pi$
  • The no. of correct decimal places from an AP: $2,5,8,\ldots$
  • The coefficient of $\pi$ term alternate as $(-1)^{n/4}4^n$
  • Other similar results and any of yours, if you observed.

Real question:

  • Can anyone put more insight to this as to explain the results?Isn't there any circular reasoning involved?
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This integral has a series equivalent

$$\frac{22}{7}-\pi=\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}$$

More generally, $$\sum_{k=m}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4m}(1-x)^4}{1+x^2}dx$$

which gives closer rational approximations to $\pi$ by adding more terms of the summation.

https://math.stackexchange.com/a/1657416/134791

A similar result is obtained changing the denominator to $\frac{1-x^8}{1-x^2}=1+x^2+x^4+x^6$.

$$ \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4 (1-x)^4} {1+x^2+x^4+x^6} dx = \frac{20 \sqrt{2}}{9}-\pi$$

Other interesting denominators are $1+x+x^2$, $\frac{1-x^6}{1-x^2}$ and $\frac{1-x^{12}}{1-x^2}$, which lead to approximations involving $\sqrt{3}$ and different exponents in the numerator.

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Its famous integral :) I too researched on this some time before.

I found this, quite useful : Integral approximations to π with nonnegative integrands.

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    $\begingroup$ While that is a very useful reference, as it stands 1) this is a link-only post, 2) this could just as well have been a comment. $\endgroup$ – Semiclassical Sep 22 '14 at 18:03
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From wolfram I get this:

Backhouse (1995) used the identity $$\begin{aligned}I_{m, n} &= \int_{0}^{1}\frac{x^{m}(1 - x)^{n}}{1 + x^{2}}\,dx\\ &= 2^{-(m + n + 1)}\sqrt{\pi}\Gamma(m + 1)\Gamma(n + 1)\times{}_{3}F_{2}\left(1, \frac{m + 1}{2}, \frac{m + 2}{2};\frac{m + n + 2}{2}, \frac{m + n + 3}{2}; -1\right)\\ &= a + b\pi + c\log 2\end{aligned}$$ for positive integers $m$ and $n$ and where $a, b$ and $c$ are rational constants to generate a number of formulas for $\pi$. In particular, if $$2m - n \equiv 0\pmod{4}$$ then $c = 0$ (Lucas 2005).

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  • $\begingroup$ I suppose $F$ is hypergeometric function and $\Gamma$ is $(n-1)!$? $\endgroup$ – RE60K Sep 23 '14 at 16:56
  • $\begingroup$ @Aditya: yes, but gamma is more than factorial. however, here because of integer argument its just a factorial. $\endgroup$ – Paramanand Singh Sep 23 '14 at 16:59

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