Formula for Natural logarithm of $\pi$

Question

Does any formula or expansion exist that gives $\ln \pi$ ?

The expansion should not just be any formula of $\pi$ with a $\ln$ before it. For example $\ln \pi$ = k + $\sum f(x)$ or something of this type.

Answer 1

One could use this formula to get

$$\ln(\pi)=\gamma+\sum_{n=1}^\infty\left[2\ln(n)-2\ln(n-0.5)-\frac1n\right]$$

Using the Euler product of the Riemann zeta function,

$$2\ln(\pi)=\ln(6)-\sum_{p\in\mathbb P}\ln(1-p^{-2})$$

where $p$ is prime.

Another interesting series:

$$\ln(\pi)=\ln(2)+2\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)$$

Answer 2

Let the matrix $A$ be the lower triangular matrix defined as: $$A=\text{If } n \bmod k=0 \text{ then } \frac{1}{n^c} \text{ else } 0$$ and let the matrix $B$ be the upper triangular matrix defined as: $$B=\text{If } k \bmod n=0 \text{ then } \frac{n \cdot \mu (n)}{k^s} \text{ else } 0$$
where $\mu(n)$ is the Möbius function.

Then the matrix product of the matrices $A$ and $B$ is the symmetric matrix $T$ starting:

$$T=A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which by analytic continuation has the Dirichlet generating function:

$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$

Then take the limit:

$$\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$

which is equivalent to subtracting the first column in the matrix $T$.

Evaluate at $s=0$ $$\frac{\zeta '(0)}{\zeta (0)}=\log(2\pi)$$

subtract with $\log(2)$

$$\frac{\zeta '(0)}{\zeta (0)}-\log(2) = \log(\pi)$$

So by analytic continuation:

$$\log(\pi)=\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T_1(n,k)}{n^1 \cdot k^0}-\log(2)$$

where

$$T_1 = \left( \begin{array}{ccccccc} +0&+1&+1&+1&+1&+1&+1&\cdots \\ +0&-1&+1&-1&+1&-1&+1 \\ +0&+1&-2&+1&+1&-2&+1 \\ +0&-1&+1&-1&+1&-1&+1 \\ +0&+1&+1&+1&-4&+1&+1 \\ +0&-1&-2&-1&+1&+2&+1 \\ +0&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

$\log(2)$ can be subtracted from the matrix as a sum from the second column:

So again by analytic continuation (not numerically directly):

$$\log(\pi)=\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T_2(n,k)}{n^1 \cdot k^0}$$

where

$$T_2 = \left( \begin{array}{ccccccc} +0&+0&+1&+1&+1&+1&+1&\cdots \\ +0&-0&+1&-1&+1&-1&+1 \\ +0&+0&-2&+1&+1&-2&+1 \\ +0&-0&+1&-1&+1&-1&+1 \\ +0&+0&+1&+1&-4&+1&+1 \\ +0&-0&-2&-1&+1&+2&+1 \\ +0&+0&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

Hope this helps.

Answer 3

The $ \sin $ function has the beautiful representation as a product \begin{eqnarray*} \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \left(1- \frac{x^{2}}{n^2 \pi^2 }\right) . \end{eqnarray*} Let $x= \frac{\pi}{2}$ and we have the Wallis product formulae for $\frac{\pi}{2}$ \begin{eqnarray*} \frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{2n}{2n-1} \frac{2n}{2n+1} . \end{eqnarray*} Taking logarithms of this will give \begin{eqnarray*} \ln \pi = \ln(2)+ \sum_{n=1}^{\infty} \left(2\ln(2n) -\ln(2n-1) -\ln(2n+1) \right) . \end{eqnarray*}

Answer 4

One has $\zeta'(0)=\log{\sqrt{2\pi}}$, and there are plenty of different ways of expressing this: for example, $$ \lim_{s \downarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}\log{n} = \log{\sqrt{2/\pi}}. $$

The series $\sum_{n=1}^{\infty} (-1)^n\log{n} $ does not converge by itself, but can be regularised in various ways: for example, if $S_m=\sum_{n=1}^{m} (-1)^n\log{n}$ we have $$ S_{2m}+S_{2m+1} = 2\sum_{n=1}^{2m} (-1)^n\log{n}-\log{(2m+1)}, $$ which can be written in closed form as $$ \log{(\pi/2)} + \log{\frac{m!}{\Gamma(m+3/2)\Gamma(m+1/2)}} $$ Stirling's formula implies that the second term converges to $0$, so $$ \log{\pi} = \log{2} + \lim_{m \to \infty} 2\sum_{n=1}^{2m} (-1)^n\log{n}-\log{(2m+1)}. $$

Answer 5

In this answer, it is shown that $$ \sum_{n=1}^\infty\frac{\zeta(2n)-1}{n+1}=\frac32-\log(\pi) $$

Answer 6

We can also derive that $$1-\sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)=ln(\pi)$$ If we use $\int_0^{\frac{\pi}{2}} ln(sin(x)) dx = -\frac{\pi}{2}ln(2)$ and replace $sin(x)$ by it's Euler product, we get $$\int_0^{\frac{\pi}{2}} ln(x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2})) dx = -\frac{\pi}{2}ln(2)$$ We can replace the product by a summation to arrive at $$\int_0^{\frac{\pi}{2}} ln(x) + \sum_{n=1}^\infty ln(1-\frac{x^2}{n^2\pi^2}) dx$$ Which equals $$\left| xln(x)-x \right|_{0}^{\frac{\pi}{2}}+\sum_{n=1}^\infty \int_0^{\frac{\pi}{2}} ln(1-\frac{x^2}{n^2\pi^2}) dx$$ We see $$\lim_{x \rightarrow 0} xln(x) = \lim_{x \rightarrow 0} \frac{\ln(x)}{\frac{1}{x}} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}} =\lim_{x \rightarrow 0} -x = 0$$ For the integral we see $$\int_0^{\frac{\pi}{2}} ln(1-\frac{x^2}{n^2\pi^2}) dx =\int_{0}^{\frac{\pi}{2}} ln(1-\frac{x}{n\pi}) + \ln(1+\frac{x}{n\pi}) dx\\ = \left|(x-\pi n)(ln(1-\frac{x}{\pi n})-1)+(x+\pi n)(ln(1+\frac{x}{\pi n})-1)\right|_0^{\frac{\pi}{2}}\\ =(\frac{\pi}{2}-\pi n)ln(1-\frac{1}{2n})+\pi n -\frac{\pi}{2}+(\frac{\pi}{2}+\pi n)ln(1+\frac{1}{2n})-\pi n -\frac{\pi}{2}\\ =\frac{\pi}{2}ln(1-\frac{1}{4n^2})+\pi n(ln(1+\frac{1}{2n})-ln(1-\frac{1}{2n}))-\pi\\ =\frac{\pi}{2}(ln(1-\frac{1}{4n^2})+4n\coth^{-1}(2n)-2)$$ So we see $$-\frac{\pi}{2}ln(2)=\frac{\pi}{2}ln(\frac{\pi}{2})-\frac{\pi}{2} +\sum_{n=1}^\infty (\frac{\pi}{2}(ln(1-\frac{1}{4n^2})+4n\coth^{-1}(2n)-2))$$ Dividing by $\frac{\pi}{2}$ and swapping some terms yields $$ln(\pi)=ln(2)+ln(\frac{\pi}{2}) = 1 - \sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)$$