The following integral,

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

is clearly positive, which proves that $\pi < 22/7$.

Is there a similar integral which proves $\pi > 333/106$?

  • Hey, how do you know that it is `clearly positive?` is there something about the integral that makes it positive? – Tyler Hilton Aug 10 '10 at 20:01
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    If $f(x)>0$ then $$\int\limits_{a}^{0} f(x)>0$$ –  Aug 10 '10 at 20:06
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    @Affan, you're integrating a fraction of even powers: it can't be negative. – Andrea Ambu Aug 11 '10 at 07:13
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    You can use $\pi = \int_0^1 \frac{4}{1+x^2}$. Now the power series of $\frac{4}{1+x^2}$ is alternating, thus stopping after odd/even numbers of terms gives under and overevaluations. Thus, for all $a< \pi – N. S. Aug 03 '12 at 13:25
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    @AndreaAmbu The powers don't have to be even: $x$ and $(1-x)$ are non-negative between $0$ and $1$. – Jaume Oliver Lafont Feb 15 '16 at 21:48

7 Answers7


This integral would do the job:

$$\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{530(1+x^2)}\:dx= \pi -\frac{333}{106}$$

  • Also you can refer to S.K. Lucas Integral proofs that $355/113 > \pi$, Gazette, Aust. Math. Soc. 32 (2005), 263-266.

  • This is the link. (Thanks to lhf for pointing out.)

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    How does one come up with that integral, may I ask? – ShreevatsaR Aug 09 '10 at 21:33
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    The link above is broken. The current one is http://educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf – lhf Jun 10 '11 at 19:05
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    @ShreevatsaR Very old thread, but for the benefit of future searchers, how one comes up with that integral along with Maple code for doing so is discussed on pages 8-9 of Lucas' writeup. – Joshua P. Swanson Feb 08 '22 at 08:17

Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the Wikipedia page about this integral. For instance, $$0 < \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$

In general, $$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx <\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx <\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$

which for $n=1$ (the integral in the question) gives slightly better bounds than just $\pi < 22/7$: $$ \frac{1}{1260} < \frac{22}{7} - \pi < \frac{1}{630}$$

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    Thank you! Doesn't that imply that pi is irrational? –  Aug 09 '10 at 21:17
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    @muad: Without checking the asymptotics of the respective numerators and denominators, I'm not sure. It doesn't follow simply from the fact that the left and right expressions go to 0: for instance 1 is rational, but you could still find sequences of rational numbers $x_n, y_n, z_n$ such that $x_n$ and $y_n$ go to 0, but $x_n < 1 - z_n < y_n$ for all $n$. – ShreevatsaR Aug 09 '10 at 21:31
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    @anon For this to work, you'd have to show that the rational approximants converged suitably fast. See Dirichlet's irrationality test. – awwalker Oct 19 '11 at 06:23

In the beginning of 2009 I was posting re similar issue at several sites, namely, at sci.math.symbolic, www.math.utexas.edu, etc.

To repeat: In Paper 1 Lucas found, by brute-force search using Maple programming, several different variants of integral identities which relate each of several first Pi convergents (described in terms of OEIS sequences as A002485(n)/A002486(n)) to Pi.

Further, in my above-mentioned postings, I conjectured the following identity below, which represents a generalization of Stephen Lucas' experimentally obtained identities between Pi and its convergents:

$$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$

$$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^m(k+(i+k)x^2)\big)/(1+x^2)\; dx$$

where integer n = 0,1,2,3,... serves as index for terms in OEIS A002485(n) and A002486(n), and {i, j, k, l, m} are some integers (to be found experimentally or otherwise), which are probably some functions of n.

The "interesting" (I think) part of my generalization conjecture is that "i" is present in both:

denominator of the coefficient in front of the integral and in the body of the integral itself

For example for $\frac{22}{7}$

$$\frac{22}{7} - \pi = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,\mathrm{d}x$$

with $n=3, i=-1, j=0, k=1, l=4, m=4$ - with regards to my above suggested generalization.

In Maple notation

i:=-1; j:=0; k:=1; l:=4; m:=4;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 22/7 - Pi

It also works for found by Lucas


formula for $\frac{333}{106}$

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

with $n=4, i=265, j=1, k=197, l=5, m=6$ -with regards to my above suggested generalization.

In Maple notation i:=265; j:=1; k:=197; l:=5; m:=6;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields Pi - 333/106

And it works for Lucas's formula for $\frac{355}{113}$

$$\frac{355}{113} - \pi = \frac{1}{3164}\int_{0}^{1}\frac{(x^8(1-x)^8(25+816x^2)}{(1+x^2)}$$

with $n=5, i=791, j=2, k=25, l=8, m=8$ -with regards to my above suggested generalization.

In Maple notation

i:=791; j:=2; k:=25; l:=8; m:=8;Int(x^m*(1-x)^l*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 355/113 - Pi

And it works as well for Lucas's formula for $\frac{103993}{33102}$

$$\pi - \frac{103993}{33102} = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(124360+77159x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=6, i= -47201, j=4, k=124360, l=14, m=12$ -with regards to my above suggested generalization.

In Maple notation

i:=-47201; j:=4; k:=124360; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields Pi - 103993/33102

And also it works Lucas's formula for $\frac{104348}{33215}$

$$\frac{104348}{33215} - \pi = \frac{1}{38544}\int_{0}^{1}\frac{x^{12}(1-x)^{12}(1349-1060x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=7, i= -2409, j=4, k=1349, l=12, m=12$ - with regards to my above suggested generalization.

In Maple notation

i:=-2409; j:=4; k:=1349; l:=12; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 104348/33215 - Pi

And it works as well for $\frac{618669248999119}{196928538206400}$

which, by the way, is not part of A002485/A002486 OEIS sequences:

$$\frac{618669248999119}{196928538206400} - \pi = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(77159+124360x^2)}{1+x^2}\,\mathrm{d}x$$

with $i= 47201, j=4, k=77159, l=14, m=12$ -with regards to my above suggested generalization.

In Maple notation

i:=47201; j:=4; k:=77159; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

618669248999119/196928538206400 - Pi

I do not have computer math resources (Mathematica, Maple, etc.) to experimentally prove or disprove it for all larger n (but see my comment below).

Best Regards, Alexander R. Povolotsky


Matt B. in his answer to my question on Mathematics Stack Exchange has provided analytical proof and improved my parametric formula by reducing the number of parameters from 5 to 4 (see Seeking proof for the formula relating Pi with its convergents).

$$(-1)^n (\pi- \frac{p_n}{q_n}) = \int_0^1 \frac{x^{\epsilon+2m'}(1-x)^{2m'}(\alpha + \beta x^2) }{(\alpha - \beta) 2 ^{m'-2} (-1)^{\epsilon}(1+x^2)}dx$$.

Below is the list of parameters in Matt B.'s formula for all cases, covered in Stephen Lucas' publications - the stuff to the right of the arrow sign is the actual Maple code, which one could copy (while in the "edit" mode) and then paste into (let say) Inverse Symbolic Calculator (it accepts Maple code) and run it there.

NB I replaced for brevity some parameter names used by Matt B: "alpha" by "a", "beta" by "b", "epsilon" by "c", "m' " by "p".

104348/33215 - Pi -> a:=1349;b:=-1060;p:=6;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)2^(p-2)((-1)^(c)*(1+x^2))),x=0...1)

Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;Int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)*(1+x^2))),x=0...1)

355/113 - Pi -> a:=25;b:=816;p:=4;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)*(1+x^2))),x=0...1)

Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;Int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)(1+x^2))),x=0...1)

22/7 - Pi -> a:=1;b:=0;p:=2;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)*(1+x^2))),x=0...1)

Obviously parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what Matt B named as "m'" (and I call it "p") : {2,3,4,4,6,6} ... Note that when "n" -> infinity, then the integral should come to 0 ...


Analysis of Thomas Baruchel calculations results (see his answer to my question in

Seeking proof for the formula relating Pi with its convergents)

led to observation that in originally supplied five parameter notation (i,j,k,l,m)

j = m/2 - 2

and correspondingly


This makes the original conjecture to depend on 4 parameters and to look like:

$$ (-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n)) =(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$

This observation confirms previous Matt B result (see previous update), which also is based on 4 parameters.

Based on his calculations results, Thomas Baruchel also found that even with 4 parameters, this formula yields infinite number of solutions for each n.

Thomas shared with me his calculations results and supplied me with quite a few of valid combinations of i,j,k,l values - so now I have a lot of experimentally found five-tuples {n,i,j,k,l}, which satisfy above parametrization, where n varies in the range from 2 to 26.

Based on this data, of course, it would be nice to find how (if at all) i,j,k,l are inter-related between each other and with "n" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "i" is strongly increasing as "n" is growing from 2 to 26).

If I didn't make a mistake RHS could be reduced (after performing integration) to:

(abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))

May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0, then we have:

Pi = (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))

Perhaps someone could programmatically check if there are any {i,j,k,l}, which would satisfy above?

Update #3:

Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0

$$\pi = \int_{0}^{1}\frac{4}{1+x^2}\,\mathrm{d}x$$

Should there be an infinite number of such cases?

see Randall's answer at math.stackexchange.com/a/2198869/28343

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  • One also could check and see that my above generalization formula also applies to identities obtained by Jaume Oliver Lafont, described in the "Following Lucas (2009)" section in http://www.oeis.org/wiki/User:Jaume_Oliver_Lafont/Constants#Integrals_involving_convergents_to_Pi – Alex Apr 08 '12 at 21:20
  • It would have been nice to have the information contained in the text `104348/33215 - Pi -> a:=....; Int(....)` rather as a table with columns $n; p_n, q_n; a, b, c, p$. The `Int(...)` is always the same (and btw. an inert `int(...)` so you have to use `eval()` to get the result. It should be defined just once as function `chk :=(a,b,c,p)->int(...)` so one can plug in the data from the table and see that it gives $\pm (\pi - p_n/q_n)$. IDK if the answer could be edited in that sense, I will maybe try to do it hoping that this won't give me flames and downvotes etc. – Max Oct 13 '20 at 21:53
  • @Max - go ahead, I don't mind. – Alex Oct 14 '20 at 16:55

Another solution is given by the integral

$$ 0 < \int_0^1 \frac{x^4(1-x)^8}{4(1+x^2)}dx = \pi -\frac{2419}{770} = \pi - \frac{333}{106}-\frac{1}{20405} $$

This proves the stricter condition $$ \pi > \frac{333}{106}+\frac{1}{20405} $$

which implies $$ \pi > \frac{333}{106} $$

Similarly, for the fourth convergent (formula (6) http://www.math.ucla.edu/~vsv/resource/general/Lucas.pdf)

$$0<\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx=\frac{3849155}{1225224}-\pi=\frac{355}{113}-\frac{5}{138450312}-\pi$$

Therefore $$\pi<\frac{355}{113}-\frac{5}{138450312}$$ and $$\pi<\frac{355}{113}$$

Even for the first convergent $$0<2\int_0^1 \frac{x(1-x)^2}{(1+x^2)}dx=\pi-3$$

so $$\pi>3$$

(See https://math.stackexchange.com/a/1618454/134791 for a proof for $3<\pi<4$ that uses this integral)

A series proof that $\pi>\frac{333}{106}$ is given by

$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106}$$

Jaume Oliver Lafont
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  • The 1st case in the above answer fits i=-1, j=2, k=1, l=4 for parametric formula $$ (-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n)) =(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$ The second case fits parameters: $$ i=-1, j=2, k=1, l=8 $$ The 3rd case fits at least the following 2 combinations $$ i=-1, j=-1, k=1, l=1 $$ $$ i=-2, j=-1, k=2, l=1 $$ $$ (abs(-1)*(-1)^2)^(-1)*Int((x^1*(1-x)^(2*(-1+2))*(1+(-1+1)*x^2))/(1+x^2),x=0...1)=(Pi -3)/2 $$ $$ (abs(-2)*(-1)^2)^(-1)*Int((x^1*(1-x)^(2*(-1+2))*(2+(-2+2)*x^2))/(1+x^2),x=0...1)=(Pi -3)/2 $$ – Alex Feb 05 '16 at 22:09
  • Might you have a source for that magnificent series? – clathratus Dec 17 '19 at 06:49
  • @clathratus , answered in your question https://math.stackexchange.com/questions/3479363/a-magnificent-series-for-pi-333-106 – Jaume Oliver Lafont Dec 21 '19 at 23:35

From the relationship


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

and $$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$

we obtain

$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(1-x)^2)}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = 106\pi-333$$


$$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$

The numerator can be shown to be nonnegative for $0\leq x\leq 1$.

In fact, we can get a smaller numerator that is still nonnegative, which leads to a closer approximation to $\pi$

$$\pi=\frac{21991}{7000} +\frac{1}{50} \int_0^1 \frac{x^4 (1 - x)^4 \left(4 - 27 x (1 - x)^2\right)}{\left(1 + x^2\right)} dx$$

Here the numerator has been adjusted to have a double zero in $(0,1)$ while not crossing the axis WA link

The resulting fraction is

$$\frac{21991}{7000} = \frac{22}{7}-\frac{9}{7000}$$

which allows for writing the double inequality


or, equivalently, an upper bound for the error in Archimedes' approximation


Jaume Oliver Lafont
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Let us consider the polynomial

$$P_n(x):=1-x^2+x^4-x^6+\cdots x^{2n-2}=\frac{x^{2n}+1}{x^2+1}.$$

We have

$$0<\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx<\int_0^1\left(\frac{x^{2n}}{0+1}\right)^2dx=\frac1{4n+1},$$ and the integral can be made as small as desired.

On another hand, the remainder of the long division of $(P_n(x)-1)^2$ by $(x^2+1)^2$ is a binomial $ax^2+b$, with $a,b$ integer. Then




we can get arbitrarily close rational approximations by a rational integral.

For example, with $P_2:=1-x^2+x^4$ we have

$$\frac{(x^4-x^2)^2}{(x^2+1)^2}=x^4-4x^2+8-\frac{12x^2+8}{(x^2+1)^2}$$ so that by integration

$$0<\frac15-\frac43+8-\frac{5\pi}2+1<\frac19$$ or



Without restrictions on the integrals, the question is virtually meaningless, because you can take any known bounds on $\pi$ (such as terms of the Gregory series), say $a<\pi<b$ and write

$$\int_0^a dx<\pi<\int_0^bdx.$$

Or if the function must be non-trivial, use any convergent squeezing of $\dfrac4{x^2+1}$ and

$$\int_0^1f(x)\,dx<\int_0^1\frac{4\,dx}{x^2+1}=\pi<\int_0^1g(x)\,dx$$ that you can make as tight as wanted.

Or better, any convergent squeezing of $\pi$, and


In my other answer, I showed constructively that that there are bracketings of $\pi$ with integrals of rational functions as tight as you want, even though they are pretty inefficient.