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# User:Jaume Oliver Lafont/Constants

## log(2)

The integral formulas in http://en.wikipedia.org/wiki/Natural_logarithm_of_2 suggest some variations:

$\log(2)=\frac{1}{2}\int_{0}^1 (1+4x)\log(1+x)dx$ verify

$\log(2)=\int_{0}^1 \log{\sqrt{\frac{1+x}{1-x}}}dx$ verify

$\log(2)=2\int_{0}^1 \left(\frac{1}{1+x^2} -\arctan{x}\right)dx$ verify

$\log(2)=\frac{1}{3}\int_{0}^1 x(13-24x^2)log(1+x)log(1-x)dx$ verify

$\log(2)=\int_{0}^1 \left(\frac{1}{2}+\log{\sqrt{1+x}}\right)dx$ verify

$\log(2)=\int_{0}^1\left(\frac{2x^2}{1+x^2}+\log(1+x^2)\right)dx$ verify

$\log(2)=\int_{0}^1 \left(1+\log{\sqrt{1-x^2}}\right)dx$ verify

$\log(2)=\frac{1}{\int_{0}^1 2^x dx}$ verify $=\frac{3}{2\int_{0}^1 4^x dx}$ verify $=\frac{7}{3\int_{0}^1 8^x dx}$ verify $=\frac{2^n-1}{n\int_{0}^1 2^{nx} dx}$ verify

$\log(2)=\frac{18107}{27720}+360\sum_{n\geq1}\frac{\binom{2n}{n}}{n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)4^n}$ verify (following 0.62 and 0.65 in [1])

### 1/2 < log(2) < 1

$\log(2) = \frac{1}{2} + \int_{0}^1 \frac{x}{(1+x)^2}dx = 1 - \int_{0}^1 \frac{1-x}{(1+x)^2}dx$ [2][3]

### 25/36 > log(2)

From

$\log(2)=\frac{25}{36}-\frac{1}{12}\int_{0}^1 \frac{x^2(1-x)^2}{(1+x)^2}dx$ verify

the following bounds are obtained for log(2)

$\frac{25}{36}-\frac{1}{360}<\log(2)<\frac{25}{36}-\frac{1}{1440}$

## π

### An integral and two corresponding (slow) series

Integral:

$\pi=\int_{0}^1 \log\frac{(1+x^2)^2}{(1+x)(1-x)^3}dx$ verify

Series:

$\frac{\pi}{2} = \sum_{k=0}^\infty \left[ \frac{1}{(4k+1)(4k+2)} + \frac{4}{(4k+2)(4k+3)}+ \frac{1}{(4k+3)(4k+4)}\right]$ verify

$= \sum_{k=0}^\infty \left[ \frac{1}{4k+1} + \frac{3}{4k+2} -\frac{3}{4k+3} -\frac{1}{4k+4} \right]$ verify

### Integrals to prove that 3 < π < 4

The following integrals have nonnegative integrand, so the inequalities hold.

$4-\pi=8\int_{0}^1 \frac{x(1-x)}{(1+x^2)^2}dx$ verify$=4\int_{0}^1 \frac{x^2}{1+x^2}dx$ verify $>0\,$

$\pi-3=2\int_{0}^1 \frac{x(1-x)^2}{1+x^2}dx > 0$ verify

The second integral is I1,2 in Lucas (2005)

Combining both results,

$3 < \pi < 4\,$

is obtained.

#### Simple bounds to prove that 3 + 1 / 12 < π < 3 + 1 / 6

Setting x=0 and x=1 in the denominator of the integral leads to the inequality

$\int_{0}^1 \frac{x(1-x)^2}{1+1}dx < \int_{0}^1 \frac{x(1-x)^2}{1+x^2}dx < \int_{0}^1 \frac{x(1-x)^2}{1+0}dx$

$\frac{1}{24} < \int_{0}^1 \frac{x(1-x)^2}{1+x^2}dx < \frac{1}{12}$

$\frac{1}{12} < 2\int_{0}^1 \frac{x(1-x)^2}{1+x^2}dx < \frac{1}{6}$

$\frac{1}{12} < \pi-3 < \frac{1}{6}$

Finally,

$3+\frac{1}{12} < \pi <3+\frac{1}{6}$

This remakes the development by Dalzell for 22 / 7 − π, now for this simpler integral related to π − 3.

### A sequence of integral representations of π

$\pi = \frac{16}{5} - \int_{0}^1 \frac{x^2(1-x)^2(1+x)^2}{1+x^2}dx$ verify

$\pi = \frac{332}{105} -\frac{1}{2}\int_{0}^1 \frac{x^2(1-x)^3(1+x)^3}{1+x^2}dx$ verify (332/105 is 'almost' the convergent 333/106 but the integrand in this related integral is sign-changing.)

$\pi = \frac{2176}{693} + \frac{1}{4}\int_{0}^1 \frac{x^4(1-x)^4(1+x)^4}{1+x^2}dx$ verify

$\pi = \frac{22912}{7293} -\frac{1}{16}\int_{0}^1 \frac{x^6(1-x)^6(1+x)^6}{1+x^2}dx$ verify

### Other integral representations of π

$\pi=\sqrt{3}\int_{0}^1 \log\frac{1-x+x^2}{(1-x)^2}dx$ verify

$\pi=4\int_{0}^1 \frac{1}{1+x^2}dx$ verify

$=8\int_{0}^1 \frac{1-x}{(1+x^2)^2}dx$ verify(compare to [31] in [4])

$=32\int_{0}^1 \frac{x^2}{(1+x^2)^3}dx$ verify

= (verify) (verify)

$=\frac{512}{3}\int_{0}^1 \frac{x^4}{(1+x^2)^5}dx$ verify

The following general integrals evaluate to the same rational multiples of π for nonnegative integer values of n

$\int_{0}^1 \frac{4x^{2n}}{(1+x^2)^{2n+1}}dx = \frac{\sqrt{\pi}4^{-n}\Gamma(n+\frac{1}{2})}{\Gamma(n+1)}$ verify

$\int_{0}^1 [x(1-x)]^{n-\frac{1}{2}}dx = \frac{\sqrt{\pi}4^{-n}\Gamma(n+\frac{1}{2})}{\Gamma(n+1)}$ verify

From (15) in [5]

$\pi=2\int_0^\frac{1}{\sqrt{2}} \frac{\sqrt{2}+2x+\sqrt{2}x^2-\sqrt{2}x^4-2x^5-\sqrt{2}x^6}{1-x^8}dx$ verify

After substituting $y=\sqrt{2}x$ and simplifying,

$\pi=4\int_0^1 \frac{dx}{2-2x+x^2}$ verify

is obtained. Although this integral is equivalent to a six-term series -using a positive basis-, it is actually simpler than (31) in Pi Formulas from Mathworld, which is equivalent to the four-term BBP series.

### Integrals involving convergents to Pi

$\pi-\frac{333}{106} = \int_0^1 \left(\frac{x^4(1-x)^8}{4(1+x^2)}+\frac{1}{20405}\right)dx$ verify

(333/106 is the third convergent to π, see A156618)

$\frac{\pi}{4}-\frac{172}{219} = \int_0^1 \left(\frac{x^4(1-x)^8}{16(1+x^2)}+\frac{1}{674520}\right)dx$ verify

(172/219 is the third convergent to π / 4, see A164924)

Following Lucas (2009)

$\pi-3 = \int_0^1 \frac{x^4(1-x)^4(89+90x^2)}{1+x^2}dx$ verify (there is also the simpler form I1,2)

$\frac{22}{7} - \pi= \int_0^1 \frac{x^8(1-x)^8(7745+7752x^2)}{28(1+x^2)}dx$ verify (as well as the form with numerator x4(1 − x)4)

$\pi-\frac{333}{106} = \int_0^1 \frac{x^8(1-x)^8(27095+26724x^2)}{1484(1+x^2)}dx$ verify

$\pi-\frac{208341}{66317}=\int_0^1 \frac{x^{16}(1-x)^{12}(191021+319754x^2)}{2059728(1+x^2)}dx$ verify

$\frac{312689}{99532}-\pi=\int_0^1 \frac{x^{9}(1-x)^{18}(160707+583718x^2)}{54145408(1+x^2)}dx$ verify

Type 3.3 equation (12) in [6] is a linear combination of integrals [7] (larger error) and [8] (smaller error).

#### An exercise

Given formulas [9] and [10] find an integral for $\pi-\frac{333}{106}$

Write $\pi-\frac{333}{106}$ as a linear combination of $\pi-\frac{47171}{15015}$ and $\frac{3849155}{1225224}-\pi$:

$\pi-\frac{333}{106}=\left(\pi-\frac{47171}{15015}\right)a+\left(\frac{3849155}{1225224}-\pi\right)b$

Split this equation into rational and transcendental parts:

$-\frac{333}{106}=-\frac{47171}{15015}a+\frac{3849155}{1225224}b$
$\pi=\pi a-\pi b\,$,

so

$1=a-b\,$.

Solve the system to get [11]:

$a=\frac{27095}{371}\,$
$b=\frac{26724}{371}\,$

Form the solution as a linear combination of the integrals

$\pi-\frac{333}{106}=a\int_0^1 \frac{x^8(1-x)^8}{4(1+x)^2}dx+b\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx =\int_0^1 \frac{x^8(1-x)^8(27095+26724x^2)}{1484(1+x^2)}dx$

and check it.

### Series involving convergents to Pi

#### Using binomial coefficients

$\pi-3\,$ [12]

$\frac{22}{7}-\pi$ [13]

$\pi-\frac{333}{106}$ [14]

$\frac{355}{113}-\pi$ [15]
$\pi-c\,$ [16] ($c-\pi\,$ [17])

(TODO: write sums for n>0 instead of n>=0)

$\frac{\pi}{2} = \sum_{n>0} \frac{2^n(3-n)}{\binom{2n}{n}}$ verify

$\pi-3=\sum_{n>0} \frac{2^n(12n-5)}{\binom{2n}{n}}$ verify

#### BBP series

##### Binary series

Adamchik and Wagon note as a curiosity ([18], page 8) that a series for $\frac{22}{7}-\pi$ can be written by choosing an appropriate value for r in the generalized BBP formula ([19]). Moreover, a series and its corresponding integral can be written for any c − π or π − c. If c = p / q,

$\pi-\frac{p}{q}=\frac{1}{8q}\sum_{k=1}^\infty \frac{1}{16^k}\left( \frac{840p-2600q}{8k+1} + \frac{-840p+2632q}{8k+2} +\frac{-420p+1316q}{8k+3} + \frac{-840p+2616q}{8k+4} +\frac{-210p+650q}{8k+5} + \frac{-210p+650q}{8k+6} +\frac{105p-329q}{8k+7}\right)$
$=\frac{1}{8q}\int_{0}^1\frac{y^8((-105p + 329q)y^2 + (420p - 1308q)y + (-210p + 650q))}{y^4-2y^3+4y-4}dy$

Setting p=3 and q=1, a series and an integral for the fractional part of π is obtained:

$\pi-3=\frac{1}{4}\sum_{k=1}^\infty\frac{1}{16^k}\left( -\frac{40}{8k+1} + \frac{56}{8k+2} +\frac{28}{8k+3} + \frac{48}{8k+4} +\frac{10}{8k+5} + \frac{10}{8k+6} -\frac{7}{8k+7}\right)$
$=\frac{1}{4}\int_{0}^1\frac{y^8(10-24y+7y^2)}{y^4-2y^3+4y-4}dy$ verify

Similarly, p=22 and q=7 yields the series pointed out by Adamchik and Wagon:

$\frac{22}{7}-\pi=\frac{1}{8}\sum_{k=1}^\infty\frac{1}{16^k}\left( -\frac{40}{8k+1} + \frac{8}{8k+2} +\frac{4}{8k+3} + \frac{24}{8k+4} +\frac{10}{8k+5} + \frac{10}{8k+6} -\frac{1}{8k+7}\right)$
$=\frac{1}{8}\int_{0}^1\frac{y^8(10-12y+7y^2)}{y^4-2y^3+4y-4}dy$ verify

For the third convergent, setting p/q=333/106 yields

$\pi-\frac{333}{106}=\frac{1}{848}\sum_{k=1}^\infty\frac{1}{16^k}\left( \frac{4120}{8k+1} - \frac{728}{8k+2} -\frac{364}{8k+3} - \frac{2424}{8k+4} -\frac{1030}{8k+5} - \frac{1030}{8k+6} +\frac{91}{8k+7}\right)$
$=-\frac{1}{848}\int_{0}^1\frac{y^8(1030-1212y+91y^2)}{y^4-2y^3+4y-4}dy$

The particular case for the fourth convergent (p=355, q=113) is:

$\frac{355}{113}-\pi=\frac{1}{452}\sum_{k=1}^\infty \frac{1}{16^k}\left( -\frac{2200}{8k+1} + \frac{392}{8k+2} +\frac{196}{8k+3} + \frac{1296}{8k+4} +\frac{550}{8k+5} + \frac{550}{8k+6} -\frac{49}{8k+7}\right)$
$=\frac{1}{452}\int_0^1 \frac{y^8(550-648y+49y^2)}{y^4-2y^3+4y-4}dy$ verify

Unfortunately, all these integrands change their sign in (0,1), so the integrals cannot be directly used as a proof that $|\pi-\frac{p}{q}|>0$

##### Slowly convergent series
###### A general formula

$\frac{\pi}{4}=\sum_{k=0}^\infty \left( \frac{r+1}{4k+1} - \frac{3r}{4k+2} + \frac{r-1}{4k+3} + \frac{r}{4k+4} \right) = \frac{r+8}{12} + \sum_{k=1}^\infty \left( \frac{r+1}{4k+1} - \frac{3r}{4k+2} + \frac{r-1}{4k+3} + \frac{r}{4k+4} \right)$ verify

###### Particular cases

Setting r=1,

$\pi-3 = 4\sum_{k=1}^\infty \left(\frac{2}{4k+1} -\frac{3}{4k+2} + \frac{1}{4k+4} \right)$ verify

For the second convergent, $r=\frac{10}{7}$ gives

$\frac{22}{7}-\pi = \frac{4}{7}\sum_{k=1}^\infty \left( -\frac{17}{4k+1} +\frac{30}{4k+2} - \frac{3}{4k+3} - \frac{10}{4k+4}\right)$ verify

$=\frac{4}{7}\int_{0}^1 \frac{x^4(10x^2+13x-17)}{(1+x)(1+x^2)}dx$ verify

This integrand is not nonnegative. (Find the corresponding integrals for other convergents and check whether they are nonnegative in (0,1) or not; try to find also nonnegative numerators for this denominator (1+x)(1+x^2) by adapting Lucas' algorithm).

For the third convergent 333/106, r=151/106 is the solution of

$\frac{333}{106}=4\frac{r+8}{12}$,

so

$\pi-\frac{333}{106}=\frac{2}{53}\sum_{k=1}^\infty \left( \frac{257}{4k+1} -\frac{453}{4k+2} + \frac{45}{4k+3} + \frac{151}{4k+4}\right)$ verify

Similarly,

$\frac{355}{113}=4\frac{r+8}{12}$

gives

$r=\frac{161}{113}$

Setting this r into the general equation and simplifying,

$\frac{355}{113}-\pi=\frac{4}{113}\sum_{k=1}^\infty \left( -\frac{274}{4k+1} +\frac{483}{4k+2} - \frac{48}{4k+3} - \frac{161}{4k+4} \right)$ verify

###### Another general formula

The general formula may be written as a function of p and q in the rational approximation c=p/q

$\pi-\frac{p}{q} = \frac{4}{q}\sum_{k=1}^\infty \left( \frac{3p-7q}{4k+1}- \frac{9p-24q}{4k+2} + \frac{3p-9q}{4k+3} + \frac{3p-8q}{4k+4} \right)$

From this formula, particular cases can be directly obtained from the target fraction p/q without computing r.

## (log(2))2

$(\log(2))^2 = \int_{0}^1 \int_{0}^1\frac{xy}{(1-xy)(2-xy)}dxdy$

$(\log(2))^2 = \int_{0}^1 \int_{0}^1\frac{2(1-2xy)}{(1+xy)(2-xy)}dxdy$ verify

## π2

$\pi^2 = \int_{0}^1 \int_{0}^1 \frac{12}{1+xy}dxdy = \int_{0}^1 \int_{0}^1 \frac{6}{1-xy}dxdy = \int_{0}^1 \int_{0}^1 \frac{8}{1-x^2y^2}dxdy$

$\sum_{n=1}^\infty \frac{1}{n^2\binom{2n}{n}} = \frac{\pi^2}{18}$

$\sum_{n=1}^\infty \frac{2^n}{n^2\binom{2n}{n}} = \frac{\pi^2}{8}$ (Lehmer, Am. Math. Monthly 92 (1985) no. 7, p. 449)

$\sum_{n=1}^\infty \frac{3^n}{n^2\binom{2n}{n}} = \frac{2\pi^2}{9}$ (Lehmer, Am. Math. Montly 92 (1985) no. 7, p. 449)

$\sum_{n=1}^\infty \frac{4^n}{n^2\binom{2n}{n}} = \frac{\pi^2}{2}$

$\sum_{n=1}^\infty \frac{5^n}{n^2\binom{2n}{n}}$ does not converge

## Catalan's constant G

$G = \int_{0}^1 \frac{4log(1+x)-log(1+x^2)}{1+x^2}dx = \frac{1}{4}\int_{0}^\infty \frac{4log(1+x)-log(1+x^2)}{1+x^2}dx$