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Timothy Chow
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[From mathworker21's comment]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

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

[From mathworker21's comment]

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

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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[From mathworker21's comment]

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