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Two integrals related to convergents of $\pi$ are given by

$$128\int_0^1 \frac{x^2(1-x)^2(x-\frac{1}{2})^2}{\sqrt{1-x^2}}dx = 106\pi - 333 + \frac{1}{5}$$

$$64\int_0^1 \frac{x(1-x)^3(x-\frac{1}{2})^2}{\sqrt{1-x^2}}dx = 355 - 113\pi + \frac{1}{5}$$

Are there similar expressions with positive integrands in small numbers that cancel the fractions $\frac{1}{5}$?

Jaume Oliver Lafont
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    A short paper on some integrals showing $355/113>\pi$ can be found here. – Semiclassical Dec 03 '24 at 22:03
  • Yes, I know, thanks. In that paper by Lucas larger numbers appear, which I believe could be lowered if another suitable form was found. – Jaume Oliver Lafont Dec 03 '24 at 22:04
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    See also this earlier question for $333/106$ instead: https://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106 – Semiclassical Dec 03 '24 at 22:06
  • I think that plugging this factor (x-1/2) or something similar will lead to a simpler result. I bumped into 113 and 106 while looking for 22/7, actually. – Jaume Oliver Lafont Dec 03 '24 at 22:07
  • Trying the root comes from this thread about the golden ratio, that leads to convergents very easily: https://math.stackexchange.com/a/2315273/134791 – Jaume Oliver Lafont Dec 03 '24 at 22:09
  • How did you find these integrals? – user Dec 03 '24 at 22:17
  • I try symmetric structures on WolframAlpha. x^n(1-x)^m has zeros at 0 and 1 so I thought another zero just in the middle, at 1/2, could have a nice effect. The power has to be even not to break the positive integrand. – Jaume Oliver Lafont Dec 03 '24 at 22:19
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    We may find some insight from the lower case $$16\int_{0}^{1} \frac{x^2(1-x)(x-\frac1{2})^2}{\sqrt{1-x^2}},\mathrm{d}x=-22+7\pi-\frac1{5}+\frac1{3}$$ – Nanayajitzuki Dec 03 '24 at 22:55
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    And those examples directly comes from these 7th order approximations $$64\int_{0}^{1} \frac{x^2(1-x)^2(x^2-4x+1)^2}{(1+x^2)^6}\frac{\mathrm{d}x}{1+x^2} = -22+7\pi+\frac1{5}$$ $$256\int_{0}^{1} \frac{x^2(1-x)^4(x^2-4x+1)^2}{(1+x^2)^6}\frac{\mathrm{d}x}{1+x^2} = -333+106\pi+\frac1{5}$$ $$64\int_{0}^{1} \frac{x(1-x)^6(x^2-4x+1)^2}{(1+x^2)^6}\frac{\mathrm{d}x}{1+x^2} =355-113\pi+\frac1{5}$$ – Nanayajitzuki Dec 03 '24 at 23:01
  • Unfortunately, the three simple linear combinations of summing two of them minus twice the other, to cancel the fraction, change the sign of the integrand in [0,1]. – Jaume Oliver Lafont Dec 03 '24 at 23:35
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    $\frac{22}{7}-\pi$ comes neat from a higher case:

    $$\frac{4}{3} \int_0^1 \frac{x^3(x-\frac{1}{2})^2(1-x)^2}{\sqrt{1-x^2}}dx =\frac{22}{7}-\pi$$

    https://www.wolframalpha.com/input?i=int_0%5E1+%284%29%28x%29%5E3%28x-1%2F2%29%5E2%281-x%29%5E2%2F3%2Fsqrt%281-x%5E2%29

     – Jaume Oliver Lafont Dec 04 '24 at 09:34
  • @Nanayajetzuki In my humble opinion, prior to (or in parallel with) the continuation of the investigation "whether there are similar expressions with positive integrands in small numbers that cancel the fractions 1/5?", may be it is worth to continue investigation whether there exist additional Pi convergents related equations, containing 1/5 fraction, and whether such equations could be generalized. – Alex Dec 07 '24 at 22:41
  • The unwanted fraction can also multiply $\pi$, such as in

    $$3\int_0^1 \frac{x^5(x-\frac{1}{2})^2(1-x)}{\sqrt{1-x^2}}dx = \frac{22}{7}-(1-\frac{1}{256})\pi$$

     – Jaume Oliver Lafont Dec 17 '24 at 23:36

2 Answers2

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Are these what you are looking for?

$$\int_{0}^{1} \frac{x^5 (1-x)^6 (197 + 462x^2)}{530(1+x^2)} \mathrm dx = \pi - \frac{333}{106}$$

$$\int_{0}^{1} \frac{x^8 (1-x)^8 (25 + 816x^2)}{3164(1+x^2)} \mathrm dx = \frac{315}{113} - \pi$$

See S.K. Lucas' Integral approximations to $\pi$ with nonnegative integrands. I hope these number are small enough. You also derived this result! (in comparison to the one's above, very small numbers). Is there any reason your own result doesn't answer your question? :) $$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$

Maxime Jaccon
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Somewhat of a cheat here:

Put $x=355/1130=71/226$ into the Maclaurin series for $\sin(x)$ and compare successive partial sums of this alternating series with $\sin(\pi/10)=(\sqrt5-1)/4$. From the $x^5/5!$ term onwards all partial sums of the series with $x=71/226$ are greater than the quadratic surd, implying $355/113>\pi$. (The proof using integer arithmetic only does not qualify as using small numbers, see the * section below.) With $x=333/1060$ in the Maclaurin series we find the opposite outcome, so $333/106<\pi$.

*Analytically, we have

$\sin(x)>x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{x^7}{5040},$

and with $x=71/226$ this gives

$\sin(x)>\dfrac{46899910085413077601}{151771284754672343040},$

and then

$4\sin^2(x)+2\sin(x)-1>\dfrac{+651269750663507719369605594928321}{5758630719020959393617218017370859110400}>0\implies \sin(x)>\dfrac{\sqrt5-1}4.$

Oscar Lanzi
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