The following integral,∫01x4(1−x)4x2+1dx=227−π

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Roy Brady · Accepted Answer

For instance,0&amp;lt;14∫01x8(1−x)81+x2 dx=π−4717115015In general,122n−1∫01x4n(1−x)4n dx&amp;lt;122n−2∫01x4n(1−x)4n1+x2 dx&amp;lt;122n−2∫01x4n(1−x)4n dxwhich for n=1 (the integral in the question) gives slightly better bounds than just π&amp;lt;22711260&amp;lt;227−π&amp;lt;1630

Abdullah Avery · Accepted Answer

Let us consider the polynomialPn(x)=1−x2+x4−x6+⋯x2n−2=x2n+1x2+1We have0&amp;lt;∫01(Pn(x)x2+1−1x2+1)2dx&amp;lt;∫01(x2n0+1)2dx=14n+1and the integral can be made as small as desired.On another hand, the remainder of the long division of (Pn(x)−1)2 by (x2+1)2 is a binomial ax2+b, with a, b integer. Then∫01(Pn(x)x2+1−1x2+1)2dx=∫01(Q(x)+ax2+b(x2+1)2)dxAs∫01ax2+b(x2+1)2dx=b−a4+b+a8πwe can get arbitrarily close rational approximations by a rational integral.For example with P2=1−x2+x4 we have(x4−x2)2(x2+1)2=x4−4x2+8−12x2+8(x2+1)2so that by integration0&amp;lt;15−43+8−5π2+1&amp;lt;19or698225&amp;lt;π&amp;lt;23675

The following integral, \(\displaystyle{\int_{{0}}^{{1}}}{\frac{{{x}^{{4}}{\left({1}-{x}\right)}^{{4}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}={\frac{{{22}}}{{{7}}}}-\pi\)

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