Formula for Natural logarithm of $\pi$ 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.

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
$$ln(\pi)=ln(2)+ln(\frac{\pi}{2}) = 1 - \sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)$$
