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 $2\pi$
$$6,\dfrac{19}{3},\dfrac{25}{4},\dfrac{44}{7},\dfrac{333}{53},\dfrac{710}{113},...$$
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?
