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.
 A: In this answer, it is shown that
$$
\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi}
$$
A: 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
A: Here is another one

$$\int_{-\infty}^{+\infty}e^{-x^2}\cos (2x^2)\mathrm dx=\sqrt{\phi \pi\over 5}$$

A: An integral uniting some favourite mathematical constants
$$\int_{-\infty}^{+\infty}\frac{t^2}{(\phi^n t)^2+(F_{2n+1}-\phi F_{2n})(\pi t^2+\zeta(3)t-e^{\gamma})^2}\mathrm dt=1$$
Where,
$\phi$; Golden ratio
$\zeta(3)$; Apery's constant
$\gamma$; Euler-Mascheroni's constant
$e$; Euler Number
$F_{n}$; Fibonacci number
and $\pi=3.14...$
A: $$
\int_0^1 \frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}
$$
A: Consider the sequence
$1,2,2,3,3,4,4,4,...$
where $a_1=1,a_{n+1}\in\{a_n,a_n+1\}$, and $a_n$ is the number of times $n$ occurs in the sequence.  Then if we assume that $a_n$ grows asymptotically as $\alpha n^\beta$, we get 
$\alpha=\phi^{1/{\phi^2}}$
$\beta=1/\phi$.
I saw this is a textbook problem on asymptotic analysis.  It turns out that for all $n$ the asymptotic expression is well within one unit of the actual $a_n$.
A: We can prove the inequalities
$$\frac{3}{2}<\frac{8}{5}<\phi<\frac{13}{8}<\frac{5}{3}$$
with representations
$$\begin{align}\phi&=\frac{3}{2}+\frac{1}{4}\int_0^1 \frac{dx}{\sqrt{4+x}}\\
\\
\phi&=\frac{8}{5}+\frac{1}{5}\int_0^1 \frac{dx}{\sqrt{121+4x}}\\
\\
\phi&=\frac{13}{8}-\frac{1}{16}\int_0^1 \frac{dx}{\sqrt{80+x}}\\
\\
\phi&=\frac{5}{3}-\frac{1}{3}\int_0^1 \frac{dx}{\sqrt{45+4x}}\\
\end{align}$$
A: $$\int_0^1\frac{\ln(1+x-x^2)}{1-x}dx=\int_0^1\frac{\ln(1+x-x^2)}xdx=2\ln^2\varphi$$
$$\int_0^1\frac{\ln(1-3x+x^2)\ln x}{x}dx=\frac85 \zeta (3)+\frac{2}{5} \pi ^2 \ln \varphi-2 i \pi  \ln^2\varphi$$
A: An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $\phi$ factor:
$$ \frac{1}{(\sqrt{\phi\sqrt{5}})e^{2\pi/5}} = 1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}}{1+\frac{e^{-10\pi}}{1+\frac{e^{-12\pi}}{\cdots}}}}}}$$

and one can then obtain a formula like:
$$ \ln \left( \sqrt{4\phi+3}-\phi^2\right) = -\frac{1}{5}\int_{e^{-2\pi}}^1 \frac{(1-t)^5(1-t^2)^5(1-t^3)^5 \dots}{(1-t^5)(1-t^{10})(1-t^{15}) \dots}\frac{dt}{t}$$
which beautifully links integrals, $e$, $\phi$ and $\pi$. It is described for instance in Golden Ratio and a Ramanujan-Type Integral.
Not very practical though to obtain $\phi$ rational approximations.
In M. D. Hirschhorn, A connection between $\pi$ and $\phi$, Fibonacci Quarterly, 2015, another asymptotic relation is:
$$ \frac{1}{\pi}=\lim_{n\to \infty} 2n {5}^{1/4}\sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}/\phi^{5n+5/2}$$
A: So you said that series are OK, so I will offer a few:
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{n!(n+1)!4^{(2n+3)}}$$
$$\phi=2\cos (\pi/5)=2\sum_{k=0}^\infty \frac{((-1)^k (\pi/5)^{2 k}}{(2k)!}$$
$$\phi=\frac{1}{2}+\frac{\sqrt{5}}{2}=\frac{1}{2}+\sum_{n=0}^\infty 4^{-n}\binom{1/2}{n}$$
A: Not exactly a series, but might also be of interest:
$$1-\frac{1}{\phi}=\frac{1}{\phi^2}=\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\dots \right)^2  \right)^2  \right)^2 \right)^2$$

$$\frac{1}{\phi^4}=\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\dots \right)^2  \right)^2  \right)^2 \right)^2$$
$$\frac{1}{\phi^4}=\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\dots \right)^2  \right)^2  \right)^2 \right)^2$$
A: Let $ F_0=0, F_1=1 ; F_{n+1}=F_{n-1}+F_n
$ be the Fibonacci numbers 
$\zeta(s)$  is the zeta function. Then:
$$
\prod_{n=1}^{\infty}\left[(-1)^{n+1}\phi F_n+(-1)^nF_{n+1}\right]^{n^{-(s+1)}}=\phi^{-\zeta(s)}
$$
A: $$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(1+x+x^2)^2+x^2}=\pi\cdot \sqrt{\frac{\phi}{5}}$$
A: For $k>0$, we have
$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty  \ln \left( \frac{x^2-2kx+k^2}{x^2+2kx\cos \sqrt{\pi^2-\phi}+k^2}\right) \;\frac{\mathrm dx}{x}=\phi}}$$
I hope you find this integral interesting.
Extra:
$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty  \frac{x^{\frac\pi5-1}}{1+x^{2\pi}} \mathrm dx=\phi}}$$
A: $$\int_{-1}^1 dx \frac1x \sqrt{\frac{1+x}{1-x}} \log{\left (\frac{2 x^2+2 x+1}{2 x^2-2 x+1}\right )}  = 4 \pi \operatorname{arccot}{\sqrt{\phi}}$$
A: Double integral of $\phi$
$$-\frac{1}{5}\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{\left(\frac{1}{5}-x+x^2\right)\sqrt{1-y+\frac{y^2}{5}}}=\ln\left(\frac{1}{\phi^4}\right)\ln\left(\frac{\phi^2+1}{\phi^4}\right)$$
Without the natural logarithm
$$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(\phi+y^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{\phi^2}$$
A: Here's a series:
$$
\phi = 1 + \sum_{n=2}^\infty \frac{(-1)^{n}}{F_nF_{n-1}}
$$
where $F_n$ is the $n$th Fibonacci number.
To see this, rewrite the numerator using the identity $(-1)^n=F_{n+1}F_{n-1}-F_n^2$, at which point the summand becomes
$$
\frac{F_{n+1}F_{n-1}-F_n^2}{F_nF_{n-1}}=\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}}
$$
and so the sum telescopes: the partial sum ending at $n$ is equal to
$$
\frac{F_{n+1}}{F_n}-\frac{F_2}{F_1}=\frac{F_{n+1}}{F_n} - 1
$$
which gives the original expression for the series via the limit $\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \phi$.
A: Based on the fact that $\varphi = \frac{1+\sqrt{5}}{2}$:

$$\varphi = \int_4^5 \frac32+\frac1{4\sqrt{x}} \mathrm{d}x$$

Based on the fact that $\varphi = 2\cos(\frac{\pi}{5})$:

$$\varphi = \int_{\tfrac{\pi}{5}}^{\tfrac{\pi}{2}} 2\sin(x) \mathrm{d}x$$

A: $$\int_0^{\infty} \frac{x^2}{1+x^{10}} \, \mathrm{d}x = \frac{\pi}{5 \phi}.$$
A: -I remember really liking this one:
$$\int_0^1 \int_0^1 \frac{\text{dx dy}}{\varphi^6-x^2y^2}=\frac{\pi^2-18\log^2\varphi}{24\varphi^3}$$
I most liked it because it was specific to $\varphi$
-Also, we can note this M.SE result (with some interpolation)
$$\int_0^1 \frac{\log (1+x^{\alpha+\sqrt{\alpha^2-1}})}{1+x}\text{dx}=$$$$\frac{\pi^2}{12}\left(\frac{\alpha}{2}+\sqrt{\alpha^2-1}\right)+\log(\varphi)\log(2)\log(\sqrt{\alpha+1}+\sqrt{\alpha-1})\log(\text{something})$$
Perhaps someone can help me fill in $\text{"something"}$
A: Here is another one
$$
\int_0^\infty \frac{1}{5^{\frac{x}{4}}+5^{\frac{1}{2}}-5^0}dx=\phi
$$
A: This one is a bit messy.
$$
\int_0^\infty \frac{1}{(\sqrt5^x)^{2^{-(\sqrt5-1)}}+\sqrt5-1}dx=2^{\phi^{-3}}\cdot\phi
$$
A: $$\int_0^\infty \frac{1}{1+x^{\frac{10}{3}}}dx=\frac{3\pi}{5\phi}$$
A: Notice that $\frac{2}{1+\sqrt5}=\frac{1}{\phi}$
$$\int_0^1\frac{2}{(1+\sqrt5x)^2}dx=\frac{1}{\phi}$$
A: $$\int_0^1  \frac{200\sqrt5(1-x^2)-300(1-x)^2}{ \left[5\sqrt5(1+x)^2-15(1-x^2)+2\sqrt5(1-x)^2  \right]^2}dx=(2\phi+1)(\phi+2)$$
A: 
$$\int_{0}^{\phi}(1-x+x^2)^{1/\phi}(1-\phi^2x+\phi^3x^2)\mathrm dx=2^{\phi}\cdot\phi$$

A bit over-crowed in term of $\phi$
A: 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
$$
\fbox{$\Large\int_0^\infty{x^{-7/10}\over1+x}\mathrm dx={2\pi\over\phi}$}
$$
A: $$\int_0^{\infty} \frac{dx}{(1+x^\phi)^\phi}=1$$
A: All the following is based on the simple fact that:
$$\phi=2 \cos \left( \frac{\pi}{5} \right)=2 \sin \left( \frac{3\pi}{10} \right)$$
These integrals are the small sample of what we can build using this identity:
$$\frac{1}{2 \pi} \int_0^{\infty} \frac{dx}{(1+x)x^{0.7}}=\phi-1$$
$$\frac{1}{1.4 \pi} \int_0^{\infty} \frac{dx}{(1+x)^2x^{0.7}}=\phi-1$$
$$\frac{1}{2 \pi} \int_0^{1} \frac{dx}{(1-x)^{0.3}x^{0.7} }=\phi-1$$
$$\frac{5}{3 \pi} \int_0^{1} \frac{x^{0.3}dx}{(1-x)^{0.3} }=\phi-1$$
$$\frac{1}{2 \pi} \int_1^{\infty} \frac{dx}{(x-1)^{0.3}x }=\phi-1$$
$$\frac{1}{0.21 \pi} \int_0^{\infty} \frac{x^{0.3}dx}{(1+x)^{3} }=\phi-1$$
Take any tables of definite integrals, find any one that ends in a trig function and set the parameters to obtain $\phi$.

You can find the following infinite product for $\phi$ here
$$2 \phi=\prod_{k=0}^{\infty}\frac{100k(k+1)+5^2}{100k(k+1)+3^2}$$
It's converging slowly, see the link for the proof using the properties of Gamma function.
By numerical computation at $50000$ terms this infinite product gives only $5$ correct digits for $\phi$, giving $1.618029$ instead of $1.618034$.
Using the infinite product for $\cos(x)$, we get:
$$\frac{\phi}{2}=\prod_{k=1}^{\infty}\left(1- \frac{4}{5^2 (2k-1)^2} \right)$$
This infinite product at $50000$ terms gives $\phi=1.618035$, only $4$ correct digits. This is actually almost the same product, because if we rearrange it we get:
$$\frac{\phi}{2}=\prod_{k=0}^{\infty}\left(\frac{100 k (k+1)+21}{100 k (k+1)+25} \right)$$
I suggest looking at this question for much more interesting product.
A: 
$$\int_0^1 \frac{2}{\left(1+\phi  x^2\right) \sqrt{1-x^2}} \, dx=\frac{\pi }{\phi }$$

A: Here is another one:
$$\int_{0}^{1}(2x-1)\left(\frac{1}{x(1-x)}+\frac{\phi\pi}{(1+x)(2-x)}+\frac{\phi \pi^2}{(2+x)(3-x)}+\frac{\phi \pi^3}{(2+x)(3-x)}+\cdots\right)-\frac{1}{\ln(1-x)}+\frac{x^{e-1}}{\ln x} \mathrm dx=1$$
A: The length of the logarithmic spiral $\rho=e^{2\theta}$ up to $\theta=0$ is given by
$$\int_{-\infty}^0\sqrt{\rho^2+\dot\rho^2}d\theta=\int_{-\infty}^0\sqrt{1+2^2}e^{2\theta}d\theta=\phi-\frac12.$$
A: How about this one:

$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\frac{2}{\phi}-\ln \phi$$

There is an infinitely nested radical in the denominator.
A finite one is also possible:

$$\int_0^{1/16} \frac{dx}{\sqrt{x+\sqrt{x}}}=\phi-2\ln (\phi)-\frac12$$

A: Potentially interesting:
$$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$
Perhaps also worthy of consideration:
$$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{\int_{-2}^2\frac{1}{1+x^2}\, dx}{\int_{-2}^2 dx}$$
A development of the first integral:
$$\log\varphi=\frac{1}{2n-1}\int_0^{\frac{F_{2n}+F_{2n-2}}{2}}\frac{dx}{\sqrt{x^2+1}}$$
$$\log\varphi=\frac{1}{2n}\int_1^{\frac{F_{2n+1}+F_{2n-1}}{2}}\frac{dx}{\sqrt{x^2-1}}$$
which stem from the relationship $(x-\varphi^m)(x-\bar\varphi^m)=x^2-(F_{m-1}+F_{m+1})x+(-1)^m$, where $\bar\varphi=\frac{-1}{\varphi}=1-\varphi$ and $F_k$ is the $k$th Fibonacci number. I particularly enjoy:
$$\log\varphi=\frac{1}{3}\int_0^{2}\frac{dx}{\sqrt{x^2+1}}$$
$$\log\varphi=\frac{1}{6}\int_1^{9}\frac{dx}{\sqrt{x^2-1}}$$
$$\log\varphi=\frac{1}{9}\int_0^{38}\frac{dx}{\sqrt{x^2+1}}$$
$$\log\varphi=\frac{1}{12}\int_1^{161}\frac{dx}{\sqrt{x^2-1}}$$
A: I am not taking credit for this. I am just posting this because it answers the question. I give Felix Marin and Olivier Oloa complete credit for these results.
$$\int_0^{\pi/2} \ln(1+4\sin^2 x)\text{ d}x=\pi\log\left(\varphi\right)$$
and 
$$\int_0^{\pi/2} \ln(1+4\sin^4 x)\text{ d}x=\pi\log \frac{\varphi+\sqrt{\varphi}}{2}$$
Again, not mine. But they definitely deserve to be here
A: Five:Pi:Phi
Coincidence closed form
$$\int_{0}^{\pi\over 2}\mathrm dx \sqrt[5]{\tan(x)}\cdot{\ln(\csc^2(x))\over \sin(2x)}=\color{blue}5\color{red}\pi\color{brown}\phi$$
A: $$\int_0^\infty x(2x-1)\,\delta(x^2-x-1)\,dx$$

Update:
As pointed by Yuriy, we must take into account the derivative of the argument of the $\delta$ function. This is why the corrective factor $2x-1$ appears.
More generally,
$$\int_I x|g'(x)|\delta(g(x))\,dx$$ evaluates to the root of $g$ contained in the interval $I$, provided there is only one. The first factor $x$ can be replaced by any function $f(x)$ to yield the value of that function at the root.
A: $$
\int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\phi\pi}{5}
$$
A: 
$$\int_{0}^{1}{1-x^{\phi}\over 1-x}+{x(1-\phi x^{1\over \phi}+{1\over \phi}x^{\phi})\over (1-x)^2}\mathrm dx=\phi$$

A: Connecting three well-known constants together:
$$\int_{-\infty}^{+\infty}\sin^2\left(\frac{x}{\phi+\frac{1}{\phi}}\right)\cdot \frac{\mathrm dx}{4x^4+5x^2}=\frac{\pi}{e}\cdot\frac{1}{\left(\phi+\frac{1}{\phi}\right)^3}$$
A: $\textrm{ Let }y=\frac{1}{1+x^{2 \pi}}, \textrm{ then }x=\left(\frac{1}{y}-1\right)^{\frac{1}{2 \pi}} \textrm{ and }d x=-\frac{1}{2 \pi}(1-y)^{\frac{1}{2 \pi}-1} y^{-\frac{1}{2 \pi}-1} d y.$
Putting back changing the integral to a Beta function
$\displaystyle \begin{aligned} \int_0^{\infty} \frac{x^{\frac{\pi-5}{5}}}{1+x^{2 \pi}} d x &=\frac{1}{2 \pi} \int_0^1 y^{-\frac{1}{10}}(1-y)^{-\frac{9}{10}} d y \\ &=\frac{1}{2 \pi} B\left(\frac{9}{10}, \frac{1}{10}\right) \end{aligned} \tag*{} $
Using the reflection property of Beta function: $B(x, 1-x)=\pi \csc(\pi x) \textrm{  for all } x \notin \mathbb{Z}$ yields
$\displaystyle \boxed{ \int_0^{\infty} \frac{x^{\frac{\pi-5}{5}}}{1+x^{2 \pi}} d x =\frac{1}{2 \pi} \pi \csc \left(\frac{\pi}{10}\right) =\frac{\sqrt{5}+1}{2}=\phi} \tag*{} $
A: 
This is my attempt to find another integral representation for Golden ratio using some special function as shown in the above image !!!!!!!
