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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?

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

Jaume Oliver Lafont
  • 5,212

1 Answers1

3

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

Dr. Sonnhard Graubner
  • 95,283
  • I don't understand how this makes the integrand positive. Can you please elaborate? – mickep May 08 '17 at 14:33
  • you must make a partial fraction decomposition and this is the result – Dr. Sonnhard Graubner May 08 '17 at 14:40
  • 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...

     – Jaume Oliver Lafont May 08 '17 at 19:49
  • (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... – Jaume Oliver Lafont May 08 '17 at 20:19
  • The partial fraction decomposition of the first two integrals is:

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

    $$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \left(x^2 - 4 x + 5 -\frac{2}{1 + x^2} - \frac{3(x^2 - \frac{4}{3}x + 1)}{1 + x^4}\right) dx$$

     – Jaume Oliver Lafont May 09 '17 at 07:12
  • The integrals involving $\pi$ are: $$\int_0^1 \frac{dx}{1+x^2} = 2\int_0^1 \frac{x}{1+x^4}dx = \frac{\pi}{4}$$ $$\sqrt{2}\int_0^1 \frac{dx}{1+x^2} = \int_0^1 \frac{1+x^2}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}$$ – Jaume Oliver Lafont May 09 '17 at 09:58