This is inspired by problem A1 on the 1968 Putnam exam which is to prove \[\int_0^1{x^4(1-x)^4\over1+x^2}dx={22\over7}-\pi\] The numerator \(x^4(1-x)^4\) is the beta integral and is very small on the interval \([0,1]\). Since \(1\leq1+x^2\leq2\) on \([0,1]\), we can bound this integral as follows: \[0<\int_0^1{x^4(1-x)^4\over1+x^2}dx<\int_0^1x^4(1-x)^4dx ={4!4!\over9!}\approx0.001587\] This tells us \[{22\over7}-{4!4!\over9!}<\pi<{22\over7}\]
If we have some choice of exponents \(a,b\), then we can integrate with polynomial division: \[\int_0^1{x^a(1-x)^b\over1+x^2}dx =\int_0^1\left(Q(x)+{R(x)\over1+x^2}\right)dx =\int_0^1 Q(x)dx + \int_0^1 {R(x)dx\over1+x^2}\] Integrating \(Q(x)\) will give us some rational number. Since division is by a degree 2 polynomial, \(R(x)=r_1x+r_0\) for some integers \(r_1,r_0\). \[\int_0^1{r_1x+r_0\over1+x^2}dx=\int_0^1{r_1xdx\over1+x^2} +\int_0^1{r_0dx\over1+x^2}\] The first integral can be solved with substitution \(y=1+x^2\) resulting in a \(\log(2)\) part. \[{r_1\over2}\int_0^1{2xdx\over1+x^2}={r_1\over2}\int_1^2{dy\over y} ={r_1\over2}\left(\log(2)-\log(1)\right)={r_1\over2}\log(2)\] The other integral is where \(\pi\) comes from \[r_0\int_0^1{dx\over1+x^2}=r_0\left(\arctan(1)-\arctan(0)\right) =r_0\arctan(1)={r_0\pi\over4}\] In order to create a useful approximation, we need to isolate one constant, so identifying conditions where \(r_1=0\) for approximating \(\pi\) and \(r_0=0\) for \(\log(2)\). Note that both cannot be zero since \(1+x^2\) has roots \(\pm i\) which \(x^a(1-x)^b\) does not, so the remainder will be nonzero.
Suppose we have \(P(x)=x^a(1-x)^b\) and divide it by \(Q(x)=1+x^2\). Then we need the remainder \(R(x)=r_1x+r_0\) to have \(r_1=0\) for an approximation to \(\pi\). A theorem that can help is: \(P(x)\equiv c\pmod{Q(x)}\) for some constant \(c\) iff \(P(r)=c\) for each root \(r\) of \(Q(x)\). For the proof, let \(P(x),Q(x)\) be general polynomials, not the ones already stated, and assume \(Q(x)\) does not have repeated roots.
\(\Rightarrow\) Suppose \(P(x)\equiv c\pmod{Q(x)}\) for a constant \(c\). Then \(Q(x)|(P(x)-c)\) so \(P(x)-c=Q(x)H(x)\) for some other polynomial \(H(x)\). Every root of \(Q(x)\) satisfies \(Q(r)=0\) so \(P(r)-c=0\) and \(P(r)=c\).
\(\Leftarrow\) Suppose \(P(r)=c\) for a constant \(c\) at each root \(r\) of \(Q(x)\). Each root \(r\) satisfies \((x-r)|Q(x)\) and \((x-r)|(P(x)-c)\) by the factor theorem. Since this is true for all roots \(r\) without repeated roots, \(Q(x)|(P(x)-c)\) which is equivalent to stating \(P(x)\equiv c\pmod{Q(x)}\).
Since \(Q(x)=(x-i)(x+i)\), we need \(P(i)=P(-i)=c\) for some constant \(c\). \[i^a(1-i)^b=(-i)^a(1+i)^b \Rightarrow \left({1-i\over1+i}\right)^b=(-1)^a\] Since powers of these quantities are periodic, it can be easier to see the required condition in polar form. \[\left({\sqrt{2}e^{-i\pi\over4}\over\sqrt{2}e^{i\pi\over4}}\right)^b =(e^{i\pi})^a\Rightarrow (e^{-i\pi\over2})^b=(e^{i\pi})^a\Rightarrow e^{i\pi(a+{b\over2})}=1\] So we need for some integer \(z\) \[a+{b\over2}=2z\Rightarrow2a+b=4z\Rightarrow2a+b\equiv0\pmod{4}\]
For \(\log(2)\), we need a condition where \(r_0=0\) so we get a remainder of \(R(x)=r_1x\). We can use the symmetry of the roots for this. By using the remainder theorem and evaluating at roots of \(Q(x)\), \(P(i)=s_1i+s_0\) and \(P(-i)=-s_1i+s_0\) for some \(s_1,s_0\). Then we can use \(P(i)+P(-i)=2s_0=0\). \[i^a(1-i)^b+(-i)^a(1+i)^b=0\Rightarrow\left({1-i\over1+i}\right)^b=-(-1)^a\] This looks similar to what we found previously. \[(e^{-i\pi\over2})^b=-(e^{i\pi})^a=e^{i\pi a+i\pi} \Rightarrow e^{i\pi(a+1+{b\over2})}=1\] And similarly for some \(z\) so the exponent is a multiple of \(2\pi\) \[a+1+{b\over2}=2z\Rightarrow2a+2+b=4z\Rightarrow2a+b\equiv2\pmod{4}\]
We can upper bound the result using the beta integral and analyze its asymptotic behavior with Stirling's approximation. \[\int_0^1{x^a(1-x)^b\over1+x^2}dx<\int_0^1x^a(1-x)^bdx ={a!b!\over(a+b+1)!}={a!b!\over(a+b+1)(a+b)!}\] \[\sim{\sqrt{2\pi a}\left({a\over e}\right)^a \sqrt{2\pi b}\left({b\over e}\right)^b \over(a+b+1)\sqrt{2\pi(a+b)}\left({a+b\over e}\right)^{a+b}} ={1\over a+b+1}\sqrt{2\pi ab\over a+b} {a^a b^b\over(a+b)^{a+b}}\] This asymptotic behavior in 2 variables becomes more clear if we restrict it to 1 variable. For example, if \(a=b\), then it becomes \[{\sqrt{\pi a}\over2a+1}{a^{2a}\over(2a)^{2a}} ={\sqrt{\pi a}\over2a+1}4^{-a}\] So we get an error of roughly \(O(4^{-a})\).
If \(b\) grows large, then we have large binomial coefficients when doing the polynomial division, so we might like to know about the error bound if we keep \(b\) fixed somewhere so the numbers in polynomial division stay small. \[{a!b!\over(a+b+1)!}={b!\over(a+1)(a+2)\ldots(a+b)(a+b+1)}\] \[={b!\over a^{b+1}(1+1/a)(1+2/a)\ldots(1+b/a)(1+(b+1)/a)} ={b!\over a^{b+1}(1+O(1/a))}\] This shows that with \(b\) fixed, the error bound decay is described by a polynomial rather than exponential function. For a practical method to compute \(\pi\), we need rapid error decay. With this method, the only way we can get rapid error decays requires both \(a,b\) to grow together, which also means the polynomial coefficients grow exponentially. More practical ways to compute \(\pi\) have both exponential error decay and small terms.