Should $\pi$ be harder than $\log(2)$?

Simple integrals relate $\pi$ to its first two convergents $3$ and $\dfrac{22}{7}$.

$$ \begin{align} \int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx&=\pi-3\\ \\ \int_0^1\frac{x^4(1-x)^4}{1+x^2}dx&=\frac{22}{7}-\pi (Dalzell) \end{align} $$

However, the following $\dfrac{333}{106}$ and $\dfrac{355}{113}$ are more involved. Is there an integral that proves $\pi > 333/106$?

In contrast, analogous integrals that evaluate to the error between $\log(2)$ and its first four convergents are easy to find.

$$ \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx &=\frac{7}{10}-\log\left(2\right) \\ \end{align} $$

Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$

Why not $2\pi$?

Comparing the list of convergents to $\pi$ $$3,\dfrac{22}{7},\dfrac{333}{106},\dfrac{355}{113},...$$

to that of $2\pi$


shows that the convergents for $2\pi$ are not simply twice the convergents for $\pi$.

The first four convergents of $\pi$ have their correspondent fractions in the list for $2\pi$, but two more approximations appear: $$\pi \approx \frac{19}{6}$$ $$\pi\approx\frac{25}{8}$$

The list of integrals for the first four convergents of $2\pi$ is simpler than that of $\pi$.

$$ \begin{align} 4\int_0^1 \dfrac{x(1-x)^2}{1+x^2}dx&=2\pi-6\\ \\ 4\int_0^1 \dfrac{x^3(1-x)^2}{1+x^2}dx&=\dfrac{19}{3}-2\pi\\ \\ \dfrac{1}{2} \int_0^1 \dfrac{x(1-x)^4(1+4x+x^2)}{1+x^2}dx&=2\pi-\dfrac{25}{4}\\ \\ 2\int_0^1 \dfrac{x^4(1-x)^4}{1+x^2}dx&=\dfrac{44}{7}-2\pi\\ \end{align} $$

The main question is:

Are there integer $m$,$n$ and a rational $q$ for the third convergent of $2\pi$? $$q\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx=2\pi-\frac{25}{4}$$ If not, how to prove it?

and a more speculative one:

What rational multiple of $\pi$ should we focus on?


It's not a definitive and complete answer at all and it's too lenghty for a comment. It's an empirically based answer.


$\displaystyle J(m,n)=\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx$

You are searching an integer relation between $J(m,n)$ and $2\pi-\dfrac{25}{4}$

Here is a program for PARI-GP to search empirically for $m,n$:


$scan(r,p)=\{$ $for(m=1,r,for(n=1,r,L=lindep([J(m,n),2*Pi-25/4]);if(abs(L[1])<p\&\&abs(L[2])<p,print(m,\text{" "},n,\text{" "},L[1],\text{" "},L[2]);return)))\}$


p=precision for integers coefficients.

To use:

start to define the precision

\p 100

(precision to compute integrals=100)


(i have tested this, no m,n found)

If m,n are found,

$L[1]J(m,n)+L[2]\times 2\pi=0$

$L[1],L[2]$ are integers.

If you replace 25/4 by 19/3 or by 44/7 this program finds an integer relation.

Bon courage dans tes recherches !


To search for integer relation between $J(m,n),\pi,1$

i mean searching for integers $a,b,c$ and integers $m,n$ such that:


here is a variant of the program above:

$scanb(r,p)=\{$ $for(m=1,r,for(n=1,r,L=lindep([J(m,n),Pi,1]);if(abs(L[1])<p\&\&abs(L[2])<p\&\&abs(L[3])<p,print(m," ",n," ",L[1]," ",L[2]," ",L[3])))) \}$

  • $\begingroup$ The output of your second version shows that there is at least one difference that comes from two integrals (is it unique?): $$\int_0^1 \frac{x^2(1-x)^4}{1+x^2}dx=2\int\frac{x^5(1-x)^2}{1+x^2}dx=\pi-\dfrac{47}{15}$$ The fraction $\dfrac{47}{15}$ is the mediant of $\dfrac{22}{7}$ and $\dfrac{25}{8}$ $$\dfrac{47}{15}=\dfrac{22+25}{8+7}$$ Subtracting both integrals, the denominator $(1+x^2)$ cancels and we get this zero relation $$\int_0^1 x^2(1-x)^2(1-2x)dx = 0$$ wolframalpha.com/input/?i=int_0%5E1+x%5E2(1-x)%5E2(1-2x)+dx $\endgroup$ – Jaume Oliver Lafont Apr 17 '17 at 4:36
  • 1
    $\begingroup$ It's easy to prove that for $n\geq 0$, $\displaystyle \int_0^1 x^n(1-x)^n(1-2x)dx=0$ just perform the change of variable $y=1-x$ and observe that $1-2x=-(1-2(1-x))$ $\endgroup$ – FDP Apr 17 '17 at 9:50

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