Problem A1 on the 1968 Putnam exam is to prove the following which shows \(\pi<{22\over7}\) \[\int_0^1{x^4(1-x)^4\over1+x^2}dx={22\over7}-\pi\] Integrals like this can be setup to evaluate to a fraction plus or minus a multiple of \(\pi\) or \(\log(2)\), leading to an approximation because the integral itself is a small positive number. More detail about that will be in another writing. For now, we will evaluate a similar looking integral that is useful to form error bounds.
The beta integral (for integers) is \[\int_0^1 x^a(1-x)^bdx = {a!b!\over(a+b+1)!}\] This writing will be an inductive proof for integers \(a,b\geq0\). More generally, the beta integral is \[\int_0^1 t^{\alpha-1} (1-t)^{\beta-1} dt = {\Gamma(\alpha)\Gamma(\beta)\over\Gamma(\alpha+\beta)}\] For the rest of this writing, let \[I_{a,b}=\int_0^1 x^a(1-x)^b dx\]
When \(a=b=0\) then \[I_{0,0}=\int_0^1 x^a(1-x)^b dx = \int_0^1 dx = {0!0!\over1!} = 1\]
Assume \(I_{a,b}={a!b!\over(a+b+1)!}\). Then \[I_{a+1,b}=\int_0^1 x^{a+1}(1-x)^bdx\] Integrate by parts with \(u=x^{a+1}\) and \(dv=(1-x)^bdx\) \[=\left[x^{a+1}{(1-x)^{b+1}\over b+1}\right]_0^1 -\int_0^1{-(1-x)^{b+1}\over b+1}(a+1)x^adx ={a+1\over b+1}\int_0^1 x^a(1-x)^{b+1}dx\] \[={a+1\over b+1}\int_0^1 x^a(1-x)^b(1-x)dx ={a+1\over b+1}\left[\int_0^1 x^a(1-x)^bdx -\int_0^1 x^{a+1}(1-x)^bdx\right]\] So we have \[I_{a+1,b}={a+1\over b+1}\left(I_{a,b}-I_{a+1,b}\right) \Rightarrow\left(1+{a+1\over b+1}\right)I_{a+1,b}={a+1\over b+1}I_{a,b}\] \[\Rightarrow I_{a+1,b}={a+1\over a+b+2}I_{a,b}= {(a+1)a!b!\over(a+b+2)(a+b+1)!}={(a+1)!b!\over(a+b+2)!}\]
Assume \(I_{a,b}={a!b!\over(a+b+1)!}\). \[I_{a,b+1}=\int_0^1 x^a(1-x)^{b+1}dx=\int_0^1 x^a(1-x)^b(1-x)dx\] \[=\int_0^1 x^a(1-x)^bdx - \int_0^1 x^{a+1}(1-x)^bdx=I_{a,b}-I_{a+1,b}\] This dependency is ok. The induction on \(a\) gives the dependency chain with fixed \(b\). The induction on \(b\) shows how \(I_{a,b+1}\) for all \(a\) depends on \(I_{a,b}\) for all \(a\). So we continue with the algebra \[I_{a,b+1}={a!b!\over(a+b+1)!}-{(a+1)!b!\over(a+b+2)!} ={(a+b+2)a!b!-(a+1)a!b!\over(a+b+2)!}\] \[={(b+1)a!b!\over(a+b+2)!}={a!(b+1)!\over(a+b+2)!}\]
We can show the symmetry with a change of variables \(y=1-x\) \[I_{a,b}=\int_0^1 x^a(1-x)^bdx=\int_1^0(1-y)^a y^b(-dy) =\int_0^1y^b(1-y)^ady=I_{b,a}\]