Could anyone try to prove that the below conjectured formula is valid for relating $\pi$ with ALL of its convergents - those, which are described in OEIS via A002485(n)/A002486(n)?

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

and in unformatted form:


where integer $n = 0,1,2,3,...$ serves as the index for terms in OEIS A002485(n) and A002486(n), and $\{i, j, k, l\}$ 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 both "$i$" and "$j$" are present in both denominator of the coefficient in front of the integral and in the body of the integral itself

At this time it could be shown that the formula under question is applicable for some first few convergents (of the A002485(n)/A002486(n) type)

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

In Maple notation

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

yields 22/7 - Pi

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

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

yields Pi - 333/106

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

In Maple notation

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

yields 355/113 - Pi

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

In Maple notation

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

yields Pi - 103993/33102

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

In Maple notation

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

yields 104348/33215 - Pi

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

In Maple notation

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

yields 618669248999119/196928538206400 - Pi

This question relates to my answer given in https://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618

Update: Recently Thomas Baruchel (see his answer at https://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents ) has conducted extensive calculations and found that even the parametric formula (with four parameters) 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 parameterization, 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).

RHS could be reduced (after performing integration - please let me know if I made a mistake in doing this) 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?

Are there any {i,j,k,l}, which would satisfy such condition?

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))

Update #2:

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 infinite number of such cases?

P.S. Per discussion with Jaume Oliver Lafont, depending on the value of the polynomial x degree in the integral body's numerator (while denominator stays to be the same "1+x^2"), the result varies from "Pi" to "log(2)" and also to "+/- (Pi - p/q)" as well as to "+/-(log(2)-p/q)", so perhaps now one could produce two distinct families of parameterization: one for Pi and the differences between Pi and its convergents and another for log(2) and the differences between log(2) and its convergents.

P.P.S. While manipulating expressions in Wolfram Cloud Development Platform and WolframAlpha I came across the following parametric identity

Sqrt[Pi] = (1/(2^j)*((k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l] + ((k + i) Gamma[7 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -l) Gamma[5 + 2 j] Gamma[1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +1/2 (3 + j) (5 + 2 j) (k + i) HypergeometricPFQRegularized[{1,7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1])) (2)

which indeed gave Sqrt[Pi] for each set of {i,j,k,l} given in above-listed cases 1), 2), 3), 4), 5), 6)

I presume that above identity (2) will yield Sqrt[Pi] for other (infinite) number of sets of {i,j,k,l}.

Is it an interesting identity?


Best Regards,

Alexander R. Povolotsky

I was notified that Maple simplifies the expression on the right hand side of (2) to sqrt(Pi). It seems to be true for arbitrary j,k,l,m.

Based on above we could consider case:

Sqrt[Pi] = (2^(5+3 j) (Gamma[5+2 j] Gamma[8+3 j] HypergeometricPFQ[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+2 Gamma[7+2 j] Gamma[6+3 j] HypergeometricPFQ[{1,7/2+j,4+j},{4+(3 j)/2,9/2+(3 j)/2},-1]))/(Gamma[5+2 j] Gamma[6+3 j] Gamma[8+3 j] (HypergeometricPFQRegularized[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+(15+11 j+2 j^2) HypergeometricPFQRegularized[{1,7/2+j,4+j},{4+(3 j)/2,(3 (3+j))/2},-1]))


  • 2
    $\begingroup$ Possibly somewhat related: mathoverflow.net/questions/67384/… $\endgroup$
    – j.c.
    Jul 10, 2014 at 13:58
  • 6
    $\begingroup$ The integral $I_{l,m}=\int_0^1 \frac{x^l(1-x)^m}{1+x^2}\,dx$ is always a linear combination of $1$, $\pi$, and $\log 2$ with rational coefficients. So you can always find $k$ and $i+k$ so that $k I_{l,m}+(i+k)I_{l+2,m}$ is $\alpha+\beta\pi$. I'm not sure how you prove that $\beta/i$ is a power of two, though. $\endgroup$
    – Kirill
    Jul 10, 2014 at 14:34
  • 2
    $\begingroup$ It does appear, though, that $x^l(1-x)^m\bmod 1+x^2$ has always zero or a power of two as its constant coefficient. $\endgroup$
    – Kirill
    Jul 10, 2014 at 14:40
  • 2
    $\begingroup$ @Kirill, $x^{\ell}(1-x)^m=(1+x^2)p(x)+ax+b$; now evaluate at $x=i$, and note that $1-i=\sqrt2e^{-i\pi/4}$. $\endgroup$ Jul 13, 2014 at 1:07
  • 7
    $\begingroup$ May I suggest, Alex, that you learn something about formatting mathematics on this site, as your latest edit is unreadable. Have a look at how the earlier parts of your question are formatted, and/or click on the "help" link on this page. $\endgroup$ Jul 24, 2014 at 0:41

1 Answer 1


We have

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

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

Now $\frac{1}{2^j }\int_0^1x^l(1-x)^{2(j+2)}\; dx $ is just some rational number so by selecting $k$ and $i$ we can take the second term to be any rational number (positive if $k$ is required to be positive).

So if we evaluate the left term and find it to be $\pm \pi$ plus a rational number, we can deduce that for any other number of the form $\pm \pi$ plus a rational number (with the same sign in the $\pm$) we can choose $k,i$ to make the integral equal that formula.

This is easy to do. If we set

$$\frac{x^{l+2} (1-x)^{2j+2}}{1+x^2} = \frac{a+bx}{x^2} + f(x)$$

for $a,b$ two rational numbers and $f(x)$ a polynomial in $x$ with rational coefficients, then

$$x^{l+2} (1-x)^{2j+4} = a+bx +(1+x)^2 f(x)$$

$$ i ^{l+2} (1-i)^{2j+4} = a + bi$$

$$a+bi = i^{l+2} (-2i)^{j+2} = 2^2 \cdot 2^j \cdot i^{l-j} $$

As long as $j$ or $l$ are congruent modulo $2$, this is $\pm 4 \cdot 2^j$, so the leftmost term is the integral of a polynomial with rational coefficients plus

$$ \pm 4 \int_0^1 \frac{1}{1+x^2} dx = \pm \pi$$ as desired.


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