<s>[INST] Is there any integral for the Golden Ratio? I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:
$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$
$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$
$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$
Is there an interesting integral* (or some series) whose result is simply $\phi$?
* Interesting integral means that things like
$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$
are not a good answer to my question.
 [/INST] Here is a collection of the series with reciprocal binomial coefficients.
$$\sum_{n=0}^\infty (-1)^n  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=\frac{4}{5} \left(1-\frac{\sqrt{5}}{5} \ln \phi \right)$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=-\frac{2\sqrt{5}}{5} \ln \phi$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=-2 \ln^2 \phi$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=\frac{4\sqrt{5}}{5} \ln \phi$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=\frac{8\sqrt{5}}{5} \ln \phi-4 \ln^2 \phi$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{n-1}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=\frac{3\sqrt{5}}{5} \ln \phi-\frac{1}{2}$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{(n-1)^2}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=1-\sqrt{5} \ln \phi+ \ln^2 \phi$$
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2(n^2-1)}  \left( \begin{matrix} 2n  \\ n \end{matrix} \right)^{-1}=4\ln^2 \phi-\frac{\sqrt{5}}{2} \ln \phi-\frac{3}{8}$$
A one with $\pi$:
$$\sum_{n=0}^\infty \left( \begin{matrix} 4n  \\ 2n \end{matrix} \right)^{-1}=\frac{16}{15}+\frac{\sqrt{3}}{27} \pi-\frac{2\sqrt{5}}{25} \ln \phi $$
Source here
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