Very curious sequence of integrals $I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2}\mathrm dx$

Solution 1:

This problem is extensively studied in the paper Integral approximations of $\pi$ with non-negative integrands. by S. K. Lucas. See page 5 for explicit formula.

Solution 2:

Well, I guess I can't compete with this research cited by Norbert, but maybe this will be of any help. One can try to expand $(1-x)^n = \sum_{k=0}^n {n \choose k}(-x)^k$, so $$\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx= \sum_{k=0}^n {n \choose k}(-1)^k\int_0^1 \frac {x^{4n+k}} {1+x^2} \,\mathrm dx$$ Then one can change variable so $$\int_0^1 \frac {x^{4n+k}} {1+x^2} \,\mathrm dx=\frac{1}{2}\int_0^1 \frac {t^{2n+\frac{k-1}{2}}} {1+t} \,\mathrm dt$$ The last integral (in the indefinite form) is: $$\int_0^1 \frac {t^{2n+\frac{k-1}{2}}} {1+t} \,\mathrm dt=\frac{2 t^{\frac{1}{2} (k+4 n+1)} \, _2F_1\left(1,\frac{1}{2} (k+4 n+1);\frac{1}{2} (k+4 n+3);-t\right)}{k+4 n+1}$$ Plugging the limits will give: $$\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx=\frac{1}{4}\sum_{k=0}^n {n \choose k}(-1)^k\left(\psi ^{(0)}\left(\frac{k}{4}+n+\frac{3}{4}\right)-\psi ^{(0)}\left(\frac{k}{4}+n+\frac{1}{4}\right)\right)$$ where $\psi ^{(0)}\left(\frac{k}{4}+n+\frac{1}{4}\right)$ is the $0$-derivative of the digamma function $\psi(z)$.