5
$\begingroup$

From integrals

$$\pi=\frac{24\sqrt{2}}{11} + \frac{8}{11} \int_0^1 \frac{x (1 - x)^2(1 + 2 \sqrt{2} x^4)}{1 + x^2 + x^4 + x^6} dx$$

and

$$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6} dx$$

the following linear combination for Ramanujan's approximation $\pi\approx\frac{9801}{2206\sqrt{2}}$ is obtained:

$$\pi=\frac{9801}{2206\sqrt{2}} - \frac{1}{8824}\int_0^1 \frac{x (1 - x)^2 (124 (1 + 2 \sqrt{2} x^4) - 5769 \sqrt{2} x^3 (1 - x)^2)}{1 + x^2 + x^4 + x^6} dx$$

This integrand is small but has sign changes in $\left(0,1\right)$, so it does not provide a direct proof that $\pi<\frac{9801}{2206\sqrt{2}}$ such as Dalzell integral for $\pi<\frac{22}{7}$.

$$\pi = \frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$

Is there an integral for $\frac{9801}{2206\sqrt{2}}-\pi$ with positive integrand?

Related questions

Why some curious almost-identities

Is there an integral that proves $\pi > 333/106$?

Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?

An integral for $2\pi+e-9$

$\endgroup$
3
$\begingroup$

HINT: write your Integrand in the form $$2\,\sqrt {2}x-4\,\sqrt {2}+{\frac {2\,\sqrt {2}+1}{{x}^{2}+1}}+{\frac {2\,{x}^{2}\sqrt {2}-2\,\sqrt {2}x-{x}^{2}+2\,\sqrt {2}+x-1}{{x}^{4}+1 }} $$

$\endgroup$
  • $\begingroup$ I don't understand how this makes the integrand positive. Can you please elaborate? $\endgroup$ – mickep May 8 '17 at 14:33
  • $\begingroup$ you must make a partial fraction decomposition and this is the result $\endgroup$ – Dr. Sonnhard Graubner May 8 '17 at 14:40
  • $\begingroup$ The two main components are $$-\frac{1}{8824}\int_0^1 \frac{\sqrt{2} \left(29341 x^4 + (62 \sqrt{2} - 23324) x^3 + (35110 - 124 \sqrt{2}) x^2 + (62 \sqrt{2} - 23324) x + 29341\right)}{x^6 + x^4 + x^2 + 1} dx = -\pi $$ with positive integrand and $$-\frac{1}{8824} \int_0^1 \sqrt{2}(-29341 + 23324 x - 5769 x^2) dx = \frac{9801}{2206\sqrt{2}}$$ with negative integrand in $(0,1)$. I understand the problem comes when adding up, because the absolute value of the positive one is not always larger than the negative... $\endgroup$ – Jaume Oliver Lafont May 8 '17 at 19:49
  • $\begingroup$ (The integrand in the answer is the one from first integral $$\pi\approx \frac{24\sqrt{2}}{11}$$ ) I am sorry not to see the solution from the hint only yet, maybe after sleeping... $\endgroup$ – Jaume Oliver Lafont May 8 '17 at 20:19
  • $\begingroup$ The partial fraction decomposition of the first two integrals is: $$\pi=\frac{24\sqrt{2}}{11} + \frac{4\sqrt{2}}{11} \int_0^1 \left(4 (x - 2) + \frac{4 + \sqrt{2}}{1 + x^2} + \frac{(4 - \sqrt{2})(x^2 - x + 1)}{1 + x^4}\right) dx$$ $$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \left(x^2 - 4 x + 5 -\frac{2}{1 + x^2} - \frac{3(x^2 - \frac{4}{3}x + 1)}{1 + x^4}\right) dx$$ $\endgroup$ – Jaume Oliver Lafont May 9 '17 at 7:12

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.