An integral which when evaluated leads to a better bound on a familiar constant. What is the bound and the constant?

Question

An integral which when evaluated leads to a better bound on a familiar constant. What is the bound and the constant?

$$ \frac{1}{3164}\int_0^1 \frac{x^8(1-x)^8\left(25 + 816x^2\right)}{1+x^2} \, \mathrm{d}x $$

Answer 1

For completion's sake (even though it's an old question) here's a solution verifying Wolfram's result posted in the comments.

Denoting $x^8(1-x)^8\left(25 + 816x^2\right)$ as $f(x)$ we get \begin{align*} f(x) &= \left(25x^8 + 816x^{10}\right)\sum_{k=0}^{8}\binom{8}{k} (-x)^k\\ &= \sum_{k=0}^{8}25\binom{8}{k} (-1)^k x^{k+8} +\sum_{k=0}^{8}816\binom{8}{k} (-1)^k x^{k+10}\\ &= \sum_{n=8}^{16}25\binom{8}{n-8} (-1)^{n} x^{n} +\sum_{n=10}^{18}816\binom{8}{n-10} (-1)^n x^{n}\\ &= \sum_{n=8}^{9}25\binom{8}{n-8} (-1)^{n} x^{n} + \sum_{n=10}^{16}\left[25\binom{8}{n-8}+816\binom{8}{n-10} \right](-1)^{n} x^{n} +\sum_{n=17}^{18}816\binom{8}{n-10} (-1)^n x^{n}\\ &=816 x^{18} - 6528 x^{17} + 22873 x^{16} - 45896 x^{15} + 57820 x^{14} - 47096 x^{13} + 24598 x^{12} - 7928 x^{11} + 1516 x^{10} - 200 x^{9} + 25 x^{8} \end{align*} remembering that $\binom{a}{b} = \frac{a!}{b! (a-b)!}$. Applying synthetic division to simplify $\frac{x^8(1-x)^8\left(25 + 816x^2\right)}{x^2+1}$ gives: $$ \begin{array}{cc} \begin{array}{rr} \\ &-1 \\ 0& \\ \\ \end{array} & \begin{array}{|rrrrrrrrrrrrrrrrr|rr} x^{16} &x^{15} &x^{14} &x^{13} &x^{12} &x^{11} &x^{10} &x^9 &x^8 &x^7 &x^6 &x^5 &x^4 &x^3 &x^2 &x^1 &x^0 \\ \hline 816 & - 6528 & 22873 & -45896 & 57820 & -47096 & 24598 & -7928 & 1516 & -200 & 25 & 0 & 0 &0 &0 &0 &0 &0 &0 \\ & & -816 & 6528&-22057 &39368&-35763 & 7728 & 11165 &200 &- 12681 & 0 & 12656 & 0 & -12656 &0& 12656&0&-12656\\ & 0 & 0 & 0 & 0& 0& 0 &0&0 & 0 & 0 & 0 & 0 & 0&0 &0&0&0\\ \hline 816 & -6528& 22057& -39368 &35763&-7728 &-11165 & -200 & 12681 & 0 &-12656 & 0 & 12656 &0 &-12656&0 & 12656 & 0 & -12656\\ \end{array} \end{array} $$ And thus $$ \frac{x^8(1-x)^8\left(25 + 816x^2\right)}{x^2+1} = 816 x^{16} - 6528 x^{15} + 22057 x^{14} - 39368 x^{13} + 35763 x^{12} - 7728 x^{11} - 11165 x^{10} - 200 x^9 + 12681 x^8 - 12656 x^6 + 12656 x^4 - 12656 x^2 + 12656- \frac{12656}{x^2 + 1} $$ So integrating term by term evaluates the integral $I =\int_0^1 \frac{x^8(1-x)^8(25 + 816x^2)}{1+x^2} \mathrm{d}x$ as: \begin{align*} I &=\frac{816}{17} - \frac{6528}{16} + \frac{22057}{15} - \frac{39368}{14} + \frac{35763}{13} - \frac{7728}{12} - \frac{11165}{11} - \frac{200}{10} + \frac{12681}{9} - \frac{12656}{7} + \frac{12656}{5} - \frac{12656}{3} + 12656- 12656\,\underbrace{\text{arctan}(1)}_{\color{blue}{\pi/4}}\\ & = 48 -408+\frac{22057}{15}-2812+2751-644-1015-20+1409-1808+ \frac{12656}{5}\left(\color{purple}{\frac33} \right)-\frac{12656}{3}\left(\color{purple}{\frac55} \right) + 12656 -3164 \pi\\ & = 10157+\frac{22057}{15}+\frac{37968}{15} -\frac{63280}{15} -3164 \pi\\ & = 10157 -\frac{3255}{15} -3164 \pi\\ & =10157 -217 -3164 \pi\\ & = 9940 - 3164 \pi \end{align*} and dividing this result by $3164$ allows us to conclude that $$ \boxed{\frac{1}{3164}\int_0^1 \frac{x^8(1-x)^8(25 + 816x^2)}{1+x^2} \mathrm{d}x =\frac{355}{113} - \pi} $$ since $\frac{9940}{3146} = \frac{355}{113}$.

Finally, since the integrand $\frac{x^8(1-x)^8(25 + 816x^2)}{1+x^2}$ is continuous and strictly positive on $(0,1)$ we can conclude that the integral is also positive on said interval. In other words, this means $I>0 \implies \frac{1}{3164}I>0 \implies \frac{355}{113} - \pi>0 \implies\boxed{ \frac{355}{113} > \pi}$. This is a very nice result, as implied by the question, since $\frac{355}{113}$ is known to be one of the best rational approximations to $\pi$.