11
$\begingroup$

I was studying the behaviour of very curious sequence of integrals $$I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx$$ which gives a very beautiful result for $n=1$; I tried to calculate for different values of $n$ but every time what I get is a $4^{n-1}$ times $\pi$ along with a fraction that in the denominator has almost a product a consecutive primes, Can we generalize this pattern? Any help would be appreciated!

Here are few calculations:

$$ I_1=22/7-\pi $$

$$ I_2=-\frac {2^2 \cdot 43\cdot 1097} {3\cdot 5\cdot 7\cdot 11 \cdot 13} +4\pi $$

$$ I_3=\frac {13\cdot 31\cdot 13912991} {3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23}-16\pi $$

$\endgroup$
0
9
+150
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Very neat. Stuff like this makes math so exciting. $\endgroup$ – Cameron Williams Nov 26 '13 at 18:36
  • $\begingroup$ Interesting! However, the question about the primes in the denominators still stands... $\endgroup$ – Bruno Joyal Nov 27 '13 at 3:11
2
$\begingroup$

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 $\psi(z)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.