Possibly the most striking proof of Archimedes's inequality $\pi < 22/7$ is an integral formula for the difference: $$ \frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, $$ where the integrand is manifestly positive. This formula is "well-known" but its origin remains somewhat mysterious. I ask:

Who discovered this integral, and in what context?

The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968). According to J.H.McKay's report in the American Math. Monthly (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky. Is one of them the discoverer, and if so which one?

The printed solution, both in the Monthly article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily." But surely there's more to be said, because this integral is a minor miracle of mathematics:

$\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) < 1/4^4 = 1/256$. A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$. Hence $\frac{22}{7} - \pi < 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$.

$\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple. When we expand $$ \frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1}, $$ the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral. [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.] When we integrate the polynomial part, we might expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number. But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-2$.

Compare this with the next such integrals $$ \int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2 $$ and $$ \int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015}, $$ which yield better but much more complicated approximations to $\log 2$ and $\pi$...

This suggests a refinement of the "in what context" part of the question:

Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)?

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    $\begingroup$ The origin of the integral seems to be addressed in the answers below (I learned it from Jon Borwein). It is implicitly generalized to simultaneous Diophantine approximations to $1$, $\log2$ and $\pi$ by M. Hata in his famous 1993 AA paper on the irrationality measure of $\pi$ (the paper by F. Beukers mentioned in one of the answers is another attempt); see also mathoverflow.net/questions/51853/… for related links. $\endgroup$ Commented Jul 11, 2011 at 13:35
  • $\begingroup$ See also mathoverflow.net/questions/175762/… $\endgroup$
    – Alex
    Commented Jul 10, 2014 at 14:20
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    $\begingroup$ The integral $$\int_0^1 \frac{(x(1-x))^4(x(1-x)^2)^n}{2^n(1+x^2)} dx$$ gives approximations to $\pi$ by excess only. After $\frac{22}{7}$, we get $\frac{377}{120}$, not much more complicated. Similarly, $$\int_0^1 \frac{(x(1-x)^2)^n}{2^{n-2}(1+x^2)} dx$$ approximates $\pi$ by default starting with $3$, $\frac{47}{15}$... $\endgroup$ Commented Apr 11, 2017 at 17:49
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    $\begingroup$ Looks interesting. ("by excess/default" $\rightarrow$ "from above/below") $\endgroup$ Commented Apr 11, 2017 at 18:52
  • $\begingroup$ Some analysis here: math.stackexchange.com/questions/2229847/… $\endgroup$ Commented Apr 13, 2017 at 9:54

7 Answers 7


Beukers mentions (http://www.staff.science.uu.nl/~beuke106/Pi-artikel.ps) that the integrals: $$\int^{1}_{0} \frac{(x - x^2)^{4n}}{(1 + x^2)} dx$$ give approximations to $\pi$ of the form $p/q$ with $\displaystyle{\left| \pi - \frac{p}{q} \right| < \frac{1}{q^{\theta}}}$ with $\theta \rightarrow \log(4)/\log(2 e^8) = 0.738\ldots$ as $n$ goes to infinity. So these are not really arithmetically significant. He also mentions that the integrals: $$J_n = \int^{1}_{0} \frac{(x - x^2)^n}{(1+x^2)^{n+1}} dx$$ give approximations with $\theta \rightarrow 0.9058\ldots$ as $n$ goes to infinity.

However, a further variation by Hata gives the integrals: $$I_n = \int^{1}_{-1} \frac{x^{2n} (1 - x^2)^{2n}}{(1 + i x)^{3n+1}} dx,$$ with $\theta \rightarrow 1.0449\ldots$, giving an irrationality measure for $\pi$ and providing "explicit" rational approximations. Notes that $I_1 = 14 \pi - 44$ gives the approximation $22/7$. (Beukers own integral proofs of the irrationality of $\pi^2$ and $\zeta(3)$ use somewhat different integrals.)

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    $\begingroup$ A PDF version of the article is available here: nieuwarchief.nl/serie5/pdf/naw5-2000-01-4-372.pdf $\endgroup$
    – j.c.
    Commented Aug 24, 2011 at 15:33
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    $\begingroup$ Your numerology looks weird: I got $\log(4)/\log(2e^8)=0.159\dots$. $\endgroup$
    – Fan Zheng
    Commented Feb 16, 2016 at 4:29
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    $\begingroup$ @FanZheng Should it read $\log(1024)/\log(4e^8)=0.738...$? $\endgroup$ Commented Apr 13, 2017 at 10:12
  • $\begingroup$ The error $\theta \rightarrow \log(4)/\log(2 e^8) = 0.738\ldots$ comes from NAW 5/1 nr. 4 december 2000 F. Beukers'article. $\endgroup$ Commented Dec 1, 2022 at 13:21

Wikipedia quotes the following source: D.P. Dalzell, On 22/7, J. London Math. Soc. 19 (1944), 133–134, and the MathSciNet review says "By the use of integral calculus the author establishes the inequalities ${\textstyle\frac{22}{7}}-{\textstyle\frac 1{1260}}>\pi>{\textstyle\frac{22}{7}}-{\textstyle\frac 1{630}}$. He then proceeds to develop a series $\pi={\textstyle\frac{22}{7}}+\sum_{n-1}^\infty a_n$, where the $a_n$'s are less in magnitude than the terms of a geometric series of ratio ${\textstyle\frac 1{1024}}$." The latter link is cited by S.K. Lucas, Approximations to $\pi$ derived from integrals with nonnegative integrands, Amer. Math. Monthly 116 (2009), no. 2, 166–172, whose abstract is as follows: "An intriguing definite integral due to Dalzell equals $22/7 - \pi$ where the integrand is nonnegative, and can be used to derive an infinite series for $\pi$. Here we extend Dalzell's results in two ways. First we look at a new family of integrals leading to series for π that converge arbitrarily fast. Then we show how integrals with nonnegative integrands can be found that equal $z - \pi$ or $\pi - z$ for any real $z$." The paper can be downloaded from the author's page here and it contains some further historical account from the 1960s.

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    $\begingroup$ Link: en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 $\endgroup$
    – Todd Trimble
    Commented Jun 9, 2011 at 22:44
  • $\begingroup$ @Michael Renardy: Unfortunately, I can't download the paper at the moment. Could you let me (and others in a similar position) know if the integral actually appears in the paper or some different proof is given there? Thanks. $\endgroup$ Commented Jun 10, 2011 at 2:46
  • $\begingroup$ I have not looked at the paper itself either, but it seems clear from the Wikipedia article that the contribution of the paper is the proof involving this integral. The inequality itself, with a different proof, was known to Archimedes. $\endgroup$ Commented Jun 10, 2011 at 10:59
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    $\begingroup$ I think Dalzell's paper does give this integral. On 9 November 2005 a non-logged-in user with an Australian IP number first cited Dalzell's 1944 paper in the Wikipedia article. I was the original creator of the Wikipedia article in 2003. Maybe it's amusing to notice that if you look at that article's discussion page, you'll see that it survived a proposal to delete it, after somewhat heated discussion here: en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion/… $\endgroup$ Commented Jun 10, 2011 at 21:32
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    $\begingroup$ jstor.org/stable/3619597 $\endgroup$ Commented Dec 15, 2012 at 4:09

Jonathan M. Borwein, David H. Bailey and Roland Girgensohn discuss this and related formulae for $\pi$ in their book "Experimentation in Mathematics" (see Section 1.1, p. 3). They claim that

The integral was apparently shown by Kurt Mahler to his students in the mid-1960s, and it had appeared in a mathematical examination at the University of Sydney in November, 1960.

They mention also a paper by Beuker who further developed the method of integral representations to obtain the irrationality estimate $$\left|\pi-\frac{p}{q}\right|\geq\frac{1}{q^{21.04...}},$$
which holds true for all integers $p$, $q$ with sufficiently large $q$. The exponent $21.04...$ is rather far from being optimal.

A draft of the book is freely available on J.M. Borwein's home page.

  • $\begingroup$ The above reference from 1944 seems to rebut the Borwein claim, if they say this (1960) to be the first appearance. $\endgroup$
    – Junkie
    Commented Jun 10, 2011 at 1:28
  • $\begingroup$ @Junkie: Right. Unfortunately, I can't access the referenced paper now. Does the integral actually appears in the paper? $\endgroup$ Commented Jun 10, 2011 at 2:42

This integral has a series counterpart

$$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$


(UPDATE Peter Bala New series for old functions https://oeis.org/A002117/a002117.pdf, 2009, formula 5.1)

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

which may be seen as the first truncation of $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$

Therefore, this series shows the following path to $\frac{22}{7}$ $$\frac{10}{3}-\frac{240}{1·2·3·5·6·7}=\frac{10}{3}-\frac{2·5!}{\frac{7!}{4}}=\frac{10}{3}-\frac{8·5!}{5!6·7}=\frac{1}{3}\left(10-\frac{4}{7}\right)=\frac{1}{3}·\frac{66}{7}=\frac{22}{7}$$

also illustrating how only $7$ remains.

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    $\begingroup$ Nice. As (partly) explained in the Stackexchange answer, this series can be derived from the integral by writing $1/(1+x^2)$ as $\sum_{k=0}^\infty (x^{4k} - x^{4k+2})$ and writing each term of the resulting series $\sum_{k=0}^\infty \int_0^1 (x-x^2)^4 (x^{4k} - x^{4k+2}) \, dx$ as the difference between two Beta integrals: $$ B(5,4k+5) - B(5,4k+7) = \frac{4! \, (4k+4)!}{(4k+9)!} - \frac{4! \, (4k+6)!}{(4k+11)!} $$ (the numerator cancels the central factor $4k+8$ of the denominator). $\endgroup$ Commented Feb 17, 2016 at 23:40
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    $\begingroup$ Imagining that also a factor $4k+12$ cancels leads to $$\frac{22}{7}-\pi=\frac{16}{21}\sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)}$$ $\endgroup$ Commented Feb 18, 2016 at 11:06
  • $\begingroup$ Equivalent series and integral are given by $$\begin{align} &\sum_{k=0}^\infty \frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\ &= \frac{1}{28}\int_{0}^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}$$ $\endgroup$ Commented Mar 29, 2017 at 4:33

Generalizations are discussed in S. K. Lucas, Integral approximations to $\pi$ with nonnegative integrands, Amer Math Monthly 116 (2009) 166-172. If you don't have access to the Monthly, you can find a preprint on Lucas' website. Lucas agrees with Wikipedia in citing Dalzell. You might also want to see Lucas' earlier paper, Integral proofs that $355/113\gt\pi$, Gazette Aust. Math. Soc. 32 (2005) 263-266.

  • $\begingroup$ see Randall's answer at math.stackexchange.com/a/2198869/28343 $\endgroup$
    – Alex
    Commented Mar 26, 2017 at 0:09

It may be covered by the articles referred to in the earlier answers, but if you integrate $$\frac{(x-x^2)^{8k+4}}{1+x^2}$$ over the unit interval (for $k$ a non-negative integer), and rewrite $(x-x^2)^{8k+4}$ as $x^{8k+4}(1+x^{2} -2x)^{4k+2}$, then rewrite $2^{4k+2}x^{12k+6}$ as $2^{4k+2}(x^{12k+6} +1) -2^{4k+2}$, you can see that you get a rational approximation to $2^{4k} \pi$ with an error less than $4^{-(8k+4)}$, where the rational approximation is the integral of a polynomial with integer coefficients over the unit interval. However the denominator is not usually so straightforward.


One can perform a small modification to the integral so that $\pi$ can approximate any specific rational number. Take $\frac{355}{113}$ as an example, we have:


For a general rational number $p$, one can always select a proper $n$ and find two rational numbers $a,b(a\cdot b>0)$ so that


Two more complicated cases are:

\begin{eqnarray} \frac{5706674932067741}{1816491048114374}-\pi&=&\int_{0}^{1}\frac{\left(x-x^{2}\right)^{40}\left(209201241800936912831+839071483456016221416x^{2}\right)}{165116704628429110269706240\left(1+x^{2}\right)}dx\\&\approx&2.3\times10^{-31}\\\pi-\frac{6134899525417045}{1952799169684491}&=&\int_{0}^{1}\frac{\left(x-x^{2}\right)^{40}\left(209201241800936912831+839071483456016221416x^{2}\right)}{165116704628429110269706240\left(1+x^{2}\right)}dx\\&\approx&4.9\times10^{-32} \end{eqnarray}


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