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It is well known that $$ \frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, $$

Given that $\frac{355}{113}$ is an excellent approximation of $\pi$, is there any known integral representation of $\frac{355}{113}-\pi$, in which the integrand is obviously non-negative?

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    $\begingroup$ S.K. Lucas, Integral proofs that $355/113>\pi$, Gazette Aust. Math. Soc. 32 (2005) 263-266. From MathSciNet: "No simple and elegant result was found." $\endgroup$
    – Gerald Edgar
    Commented Jun 25, 2022 at 20:08
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    $\begingroup$ My favorite from the paper Gerald linked is $$\frac{355}{113}-\pi = \int_0^1 \frac{x^8(1-x)^8(25+816x^2)}{3164(1+x^2)}dx.$$ $\endgroup$
    – mathworker21
    Commented Jun 25, 2022 at 20:13
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    $\begingroup$ @mathworker21 ... that is also quoted at math.stackexchange.com/q/860499/44 $\endgroup$
    – Gerald Edgar
    Commented Jun 25, 2022 at 20:17
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    $\begingroup$ $355/113$ is not just an "excellent approximation"; it is, like $22/7,$ one of the convergents in the continued fraction representation of $\pi.$ And it differse from $\pi$ by less than the reciprocal of the square of the denominator; thus by less than $1/113^2. \qquad$ $\endgroup$
    – Michael Hardy
    Commented Jun 26, 2022 at 5:30
  • $\begingroup$ Link to a pdf of the article @Gerald cited math.ucla.edu/~vsv/resource/general/Lucas.pdf $\endgroup$
    – David Roberts ♦
    Commented Jun 26, 2022 at 8:49

1 Answer 1

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As mentioned in the comments by Gerald Edgar and mathworker21, some formulas are given in a paper by S. K. Lucas, Integral proofs that $355/113>π$, Gazette Aust. Math. Soc. 32 (2005), 263–266. (See also the author's 2009 Amer. Math. Monthly paper, Approximations to $π$ Derived from Integrals with Nonnegative Integrands.) One such formula is

$$\frac{355}{113} - \pi = \int_0^1 \frac{x^8(1-x)^8(25+816x^2)}{3164(1+x^2)}\, dx.$$

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    $\begingroup$ No doubt your intention was to bring deserved attention to mathworker21's comment, but I think it would be more appropriate to first suggest to him or to Gerald Edgar to make it into an answer. $\endgroup$
    – Yaakov Baruch
    Commented Jun 25, 2022 at 21:15
  • $\begingroup$ The moderators have now made this answer community wiki, which should address the issue that the author of the answer is different from the author of the comment. $\endgroup$
    – Timothy Chow
    Commented Jun 26, 2022 at 12:39

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