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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:

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

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

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

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

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

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

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

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

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

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

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

### 25/36 > log(2)

From

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

the following bounds are obtained for log(2)

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

## ${\displaystyle \pi }$

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

Integral:

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

Series:

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

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

### Integrals to prove that ${\displaystyle 3<\pi <4}$

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

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

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

The second integral is ${\displaystyle I_{1,2}}$ in Lucas (2005)

Combining both results,

${\displaystyle 3<\pi <4\,}$

is obtained.

#### Simple bounds to prove that ${\displaystyle 3+1/12<\pi <3+1/6}$

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

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

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

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

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

Finally,

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

This remakes the development by Dalzell for ${\displaystyle 22/7-\pi }$, now for this simpler integral related to ${\displaystyle \pi -3}$.

### A sequence of integral representations of ${\displaystyle \pi }$

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

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

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

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

### Other integral representations of ${\displaystyle \pi }$

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

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

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

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

${\displaystyle =}$ (verify) (verify)

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

The following general integrals evaluate to the same rational multiples of ${\displaystyle \pi }$ for nonnegative integer values of n

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

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

From (15) in [5]

${\displaystyle \pi =2\int _{0}^{\frac {1}{\sqrt {2}}}{\frac {{\sqrt {2}}+2x+{\sqrt {2}}x^{2}-{\sqrt {2}}x^{4}-2x^{5}-{\sqrt {2}}x^{6}}{1-x^{8}}}dx}$ verify

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

${\displaystyle \pi =4\int _{0}^{1}{\frac {dx}{2-2x+x^{2}}}}$ 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

${\displaystyle \pi -{\frac {333}{106}}=\int _{0}^{1}\left({\frac {x^{4}(1-x)^{8}}{4(1+x^{2})}}+{\frac {1}{20405}}\right)dx}$ verify

(333/106 is the third convergent to ${\displaystyle \pi }$, see A156618)

${\displaystyle {\frac {\pi }{4}}-{\frac {172}{219}}=\int _{0}^{1}\left({\frac {x^{4}(1-x)^{8}}{16(1+x^{2})}}+{\frac {1}{674520}}\right)dx}$ verify

(172/219 is the third convergent to ${\displaystyle \pi /4}$, see A164924)

Following Lucas (2009)

${\displaystyle \pi -3=\int _{0}^{1}{\frac {x^{4}(1-x)^{4}(89+90x^{2})}{1+x^{2}}}dx}$ verify (there is also the simpler form ${\displaystyle I_{1,2}}$)

${\displaystyle {\frac {22}{7}}-\pi =\int _{0}^{1}{\frac {x^{8}(1-x)^{8}(7745+7752x^{2})}{28(1+x^{2})}}dx}$ verify (as well as the form with numerator ${\displaystyle x^{4}(1-x)^{4}}$)

${\displaystyle \pi -{\frac {333}{106}}=\int _{0}^{1}{\frac {x^{8}(1-x)^{8}(27095+26724x^{2})}{1484(1+x^{2})}}dx}$ verify

${\displaystyle \pi -{\frac {208341}{66317}}=\int _{0}^{1}{\frac {x^{16}(1-x)^{12}(191021+319754x^{2})}{2059728(1+x^{2})}}dx}$ verify

${\displaystyle {\frac {312689}{99532}}-\pi =\int _{0}^{1}{\frac {x^{9}(1-x)^{18}(160707+583718x^{2})}{54145408(1+x^{2})}}dx}$ 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 ${\displaystyle \pi -{\frac {333}{106}}}$

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

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

Split this equation into rational and transcendental parts:

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

so

${\displaystyle 1=a-b\,}$.

Solve the system to get [11]:

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

Form the solution as a linear combination of the integrals

${\displaystyle \pi -{\frac {333}{106}}=a\int _{0}^{1}{\frac {x^{8}(1-x)^{8}}{4(1+x)^{2}}}dx+b\int _{0}^{1}{\frac {x^{10}(1-x)^{8}}{4(1+x^{2})}}dx=\int _{0}^{1}{\frac {x^{8}(1-x)^{8}(27095+26724x^{2})}{1484(1+x^{2})}}dx}$

and check it.

### Series involving convergents to Pi

#### Using binomial coefficients

${\displaystyle \pi -3\,}$ [12]

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

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

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

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

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

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

#### BBP series

##### Binary series

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

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

Setting p=3 and q=1, a series and an integral for the fractional part of ${\displaystyle \pi }$ is obtained:

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

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

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

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

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

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

${\displaystyle {\frac {355}{113}}-\pi ={\frac {1}{452}}\sum _{k=1}^{\infty }{\frac {1}{16^{k}}}\left(-{\frac {2200}{8k+1}}+{\frac {392}{8k+2}}+{\frac {196}{8k+3}}+{\frac {1296}{8k+4}}+{\frac {550}{8k+5}}+{\frac {550}{8k+6}}-{\frac {49}{8k+7}}\right)}$
${\displaystyle ={\frac {1}{452}}\int _{0}^{1}{\frac {y^{8}(550-648y+49y^{2})}{y^{4}-2y^{3}+4y-4}}dy}$ verify

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

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

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

###### Particular cases

Setting r=1,

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

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

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

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

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

so

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

Similarly,

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

gives

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

Setting this ${\displaystyle r}$ into the general equation and simplifying,

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

###### Another general formula

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

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

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

## ${\displaystyle (\log(2))^{2}}$

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

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

## ${\displaystyle \pi ^{2}}$

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}{\binom {2n}{n}}}}={\frac {\pi ^{2}}{18}}}$

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

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {4^{n}}{n^{2}{\binom {2n}{n}}}}={\frac {\pi ^{2}}{2}}}$

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

## Catalan's constant G

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