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.

By Euler's reflection formula, it follows that
$$
\int_0^\infty{x^{s-1}\over1+x}\mathrm dx={\pi s\over\sin(\pi s)}\tag1
$$
Accordingly, we can find an $s$ such that $\sin(\pi s)$ can be associated with $\phi$. As it turns out, we do have some special angle that allows us to do so.

By observing the geometric properties of this triangle, we can deduce the following relationship
$$
\triangle ABC\sim\triangle BDA\cong\triangle DBA
$$
which implies
$$
{BC\over AB}={AB\over BD}
$$
Now, due to the properties of isosceles triangles, we get
$$
AB=AD=CD\Rightarrow BC=BD+CD=BD+AB
$$
Thus, we obtain
$$
1+{BD\over AB}={AB\over BD}
$$
To convenience the derivation, set $AB=1,BD=y$ so that the above identity becomes
$$
1+y=\frac1y\Rightarrow y^2+y-1=0\Rightarrow y={-1+\sqrt5\over2}=\frac1\phi
$$
Again, by the properties of isosceles, we deduce
$$
CE={1+y\over2}
$$
As a result, we obtain $\cos36^\circ$ from its definition:
$$
\cos36^\circ={CE\over CD}={CE\over AB}={1+y\over2}={1+\sqrt5\over4}=\frac\phi2
$$
Now, due to the conversion that
$$
90^\circ-36^\circ=54^\circ={3\pi\over10}
$$
we obtain
$$
\sin\left(3\pi\over10\right)=\frac\phi2
$$
Therefore, setting $s=3/10$ in (1), we obtain
$$
\fbox{$\Large\int_0^\infty{x^{-7/10}\over1+x}\mathrm dx={2\pi\over\phi}$}
$$
