# 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.

• You can skim this page, on WolframAlpha; e.g. Eq (12) and (13). Commented Feb 14, 2016 at 3:15
• Related question introducing an infinite product for GR. And this question Commented Feb 14, 2016 at 3:32
• Also this. Somewhat famous locally :-) Commented Feb 14, 2016 at 9:45
• In principle, any infinite sum can be expressed as an appropriate contour integral; thus, any of the known infinite sums for $\phi$ can be expressed as contour integrals. Commented Feb 15, 2016 at 14:31
• We have the following series representation: $$\phi=\frac{1}{2}+\frac{1331}{250} \sum \limits_{n=0}^{\infty} \frac{(2n+1)!}{5^{3n+1}(n!)^2}.$$ Commented Nov 13, 2019 at 18:26

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:

$$\log\varphi=\frac{1}{3}\int_0^{2}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{6}\int_1^{9}\frac{dx}{\sqrt{x^2-1}}$$ $$\log\varphi=\frac{1}{9}\int_0^{38}\frac{dx}{\sqrt{x^2+1}}$$ $$\log\varphi=\frac{1}{12}\int_1^{161}\frac{dx}{\sqrt{x^2-1}}$$

• Wow. Did you come up with this by yourself ? Commented Feb 14, 2016 at 4:25
• @user230452 Unfortunately not! Stems from the fact that $\text{arcsinh}{\frac{1}{2}}=\log\varphi$, and this connection comes by noting that $x^2-x-1=0\implies \frac{x-\frac{1}{x}}{2}=\frac{1}{2}$
– πr8
Commented Feb 14, 2016 at 4:28
• What about $$\int_0^{1/2}\left(\frac{x}{\sqrt{x^2+1}}+3\right)\,dx$$
– user65203
Commented Feb 14, 2016 at 18:02
• +1 for the understatement, the neat answer and the awesome username. I assume you greet other $\pi r8$s by saying "$Ar^k$" for some $k\geq2$. Commented Feb 14, 2016 at 18:23
• @DavidRicherby Indeed - though I'm humbled enough by the reception this first integral seems to have received that I might be well-advised to go by $\varphi$r$8$ from here onwards ^^.
– πr8
Commented Feb 16, 2016 at 14:00

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

• Awesome!! A strict link between $\pi$ and $\phi$, I love those things. Thank you! Commented Feb 14, 2016 at 14:35
• Brilliant!! Absolutely amazing Commented Feb 14, 2016 at 15:04
• wow! this is incredible Commented Feb 14, 2016 at 16:45
• So we know $\pi=2e\int_0^{\infty}{\cos(x)\over x^2+1}\text{d}x$ and $e=\sum_{k=0}^{\infty}{1\over k!}$ from the OP, then this answer says $\int_0^\infty{\sqrt{x}\over x^2+2x+5}\text{d}x={\pi\over 2\sqrt{\Phi}}$. My immediate thought was to combine the above to get $\Phi=\left({\sum_{k=0}^{\infty}{1\over k!}\int_0^{\infty}{\cos(x)\over x^2+1}\text{d}x \over \int_0^\infty{\sqrt{x}\over x^2+2x+5}\text{d}x}\right)^2$, which might be considered "interesting". Commented Feb 14, 2016 at 22:36
• So very nice ! Somehow you perhaps can rope in $e$ too. Commented Feb 15, 2016 at 15:18

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:

$$\frac{1}{\pi}=\lim_{n\to \infty} 2n {5}^{1/4}\sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}/\phi^{5n+5/2}$$

• The genius of Ramanujan will always remain a mystery.. what a genius. Commented Feb 15, 2016 at 14:33
• And I believe it is a good thing that this remains a mystery. Commented Feb 15, 2016 at 14:58
• mind... blown... Commented Feb 18, 2016 at 5:03
• @FourierTransform right! Commented Oct 8, 2016 at 9:14
• @VladimirReshetnikov I even made a question on this (here) :D Commented Sep 1, 2018 at 18:39

For $k>0$, we have

Extra: $$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty \frac{x^{\frac\pi5-1}}{1+x^{2\pi}} \mathrm dx=\phi}}$$

• Click the box for the proof Commented Jun 1, 2016 at 8:41
• I accepted your answer because you definitely wrote me what I wanted. Thank you so much for your beautiful elegant answer. Commented Sep 10, 2016 at 10:28
• It was a bit deflating to realize that, more generally, $$\int_0^\infty \frac{x^{\pi/k-1}}{1+x^{2\pi}}\mathrm dx=\frac{1}{2}\csc\Big( \frac{\pi}{2k}\Big)$$ and the second result was just the case $k=5$. Commented Nov 12, 2016 at 16:15
• @TitoPiezasIII Everything is a special case of something. Commented Jan 3, 2017 at 13:41
• @SimplyBeautifulArt I thought something was the special case of everything? Commented Mar 20, 2019 at 16:40

$$\int_{-1}^1 dx \frac1x \sqrt{\frac{1+x}{1-x}} \log{\left (\frac{2 x^2+2 x+1}{2 x^2-2 x+1}\right )} = 4 \pi \operatorname{arccot}{\sqrt{\phi}}$$

• There is a sign error in the log term
– Teoc
Commented Feb 14, 2016 at 16:37
• @LaplacianFourier: Thanks. Commented Feb 14, 2016 at 16:37
• Ah yes, isn't this like the most upvoted post on this site? Always fun reading even though I don't know enough math do do it.. Commented Feb 14, 2016 at 21:48
• @FarazMasroor: Actually, I think the 7th or 8th-most upvoted post. But thanks - if you want to learn feel free to ask questions! Commented Feb 14, 2016 at 21:51
• ...might as well include a link: MSE 562964 Commented Feb 18, 2016 at 7:27

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 $\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \phi$.

• Was this the first definition of golden ratio or did it have a definition before that ? Commented Feb 14, 2016 at 4:27
• @user230452 $\phi = \frac { 1+ \sqrt 5}2$
– Ant
Commented Feb 14, 2016 at 9:43
• I mean, didn't that number come from the Fibonacci series itself or did it already have a definition and was found again in the Fibonacci series ? Commented Feb 14, 2016 at 10:23
• @user230452 The golden ratio first arose as a ratio between quantities $a,b$ for which ratio $a:b$ is the same as ratio $a+b:a$. Hence, it came before (or at least independently of) Fibonacci numbers. Commented Feb 14, 2016 at 10:32
• @Wojowu My point was just that the Fibonacci sequence is all about situations where you're dealing with $a$, $b$ and $a+b$, and so is the classical definition of the golden ratio. Whereas, for example, $(1+\sqrt{5})/2$ is a completely different way of defining the same number. Anyway, I'm just nit-picking. Commented Feb 14, 2016 at 21:26

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 $\varphi = 2\cos(\frac{\pi}{5})$:

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

• I wanted to do that at first, but thought it wasn't 'interesting' by OP's standards Commented Feb 14, 2016 at 12:31
• @YuriyS I just took 'not interesting' as 'directly containing $\varphi$, or a trivial variation on it' Commented Feb 14, 2016 at 12:35
• Awesome, the second one is great!! Commented Feb 15, 2016 at 14:34

$$\int_0^{\infty} \frac{x^2}{1+x^{10}} \, \mathrm{d}x = \frac{\pi}{5 \phi}.$$

• Great! Another integral that relates two constants! Thank you! Commented Feb 14, 2016 at 19:37
• @KimPeek, there is an infinite number of integrals of this kind Commented Feb 14, 2016 at 19:42
• @YuriyS The more I see, the happier I am :D Commented Feb 14, 2016 at 20:06
• Although it adds nothing, I think having $5x^2$ on the left instead of $5$ at the denominator on the right looks even prettier (if possible) Commented Feb 15, 2016 at 7:14

$$\int_0^{\infty} \frac{dx}{(1+x^\phi)^\phi}=1$$

• Astounding beauty Commented Apr 24, 2016 at 11:04
• Another integral involving $\phi$ that might be surprising at first sight :) $$\int_0^\infty\frac1{1+x^2}\frac1{1+x^\phi}dx=\frac\pi4.$$ Commented May 27, 2016 at 19:27
• @VladimirReshetnikov How would you solve both of your integrals one? My mind's blank, and I'm not able to simplify them adequately. :/ Commented Jun 17, 2017 at 4:55
• $\displaystyle\int\frac{dx}{(1+x^\phi)^\phi}=x \, (1+x^\phi)^{1-\phi}\color{gray}{+C}$, that can be checked by differentiation. The second one is really easy. Hint: if you replace $\phi$ with $\phi^2$, it will still have the same value. Commented Jun 17, 2017 at 21:34

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.

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

There is an infinitely nested radical in the denominator.

A finite one is also possible:

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

• The first one is AMAZING!! Thank you for having shared it! :O Commented Apr 11, 2016 at 16:57
• Might help in the second to note that $\ln(\phi+1)=2\ln\phi$
– πr8
Commented May 3, 2016 at 17:36
• @TimeMaster The 1st integral is only a fancy representation of $$\int_0^1 \frac{2}{1+\sqrt{1+4x}}\ dx$$ Commented Jun 10, 2016 at 7:00
• @SophieAgnesi, my secret is revealed! Curses! Commented Jun 10, 2016 at 7:55

The length of the logarithmic spiral $\rho=e^{2\theta}$ up to $\theta=0$ is given by

$$\int_{-\infty}^0\sqrt{\rho^2+\dot\rho^2}d\theta=\int_{-\infty}^0\sqrt{1+2^2}e^{2\theta}d\theta=\phi-\frac12.$$

• Nice. Can you re-adjust the spiral so that length is $\phi$ exactly ? Commented Feb 15, 2016 at 15:11
• @Narasimham: I don't see an immediate way to achieve that.
– user65203
Commented Feb 15, 2016 at 15:22
• You already have $\sqrt{5}$ under your integral. Good example though Commented Feb 23, 2016 at 23:16

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

• Beautiful! Thank you for having posted them here. The first one is so beautiful!! Commented Apr 2, 2016 at 13:42

Five:Pi:Phi

Coincidence closed form

$$\int_{0}^{\pi\over 2}\mathrm dx \sqrt[5]{\tan(x)}\cdot{\ln(\csc^2(x))\over \sin(2x)}=\color{blue}5\color{red}\pi\color{brown}\phi$$

$$\int_0^\infty x(2x-1)\,\delta(x^2-x-1)\,dx$$

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.

• Beautiful!! Dirac Delta. Very easy and elegant, thank you! Commented Feb 14, 2016 at 17:43
• A great idea, actually! We can do it for any algebraic number, it seems Commented Feb 14, 2016 at 19:24
• @YuriyS: yep, provided you isolate the desired root in an interval.
– user65203
Commented Feb 14, 2016 at 19:36
• Actually, Wolframalpha gives another value for this integral: wolframalpha.com/input/… Commented Feb 14, 2016 at 20:19
• In general $\delta [g(x)]=\sum_k \frac{\delta (x-x_k)}{| g'(x_k)|}$ Commented Feb 14, 2016 at 20:25

$$\int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\phi\pi}{5}$$

• Awesome one!!!! Commented May 4, 2016 at 9:27

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 $$\pi=3.14...$$

• This is awesome! Commented Feb 10, 2019 at 21:37

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

Here is another one

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

• Very cool one!! Commented Apr 3, 2017 at 7:54

$$\int_0^1 \frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}$$

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 $a_n$.

• I give up! How do I put braces around an explicitly written set!?! Commented Apr 29, 2016 at 10:46
• Use \{ and \} instead of the normal braces. Commented May 2, 2016 at 13:57

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}

$$\int_0^1\frac{\ln(1+x-x^2)}{1-x}dx=\int_0^1\frac{\ln(1+x-x^2)}xdx=2\ln^2\varphi$$ $$\int_0^1\frac{\ln(1-3x+x^2)\ln x}{x}dx=\frac85 \zeta (3)+\frac{2}{5} \pi ^2 \ln \varphi-2 i \pi \ln^2\varphi$$

• Could those people vote to delete please explain the reason? Commented Oct 27, 2018 at 13:02
• Maybe they are just ignorant. Commented Sep 1, 2020 at 11:14

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(1+x+x^2)^2+x^2}=\pi\cdot \sqrt{\frac{\phi}{5}}$$

So you said that series are OK, so I will offer a few:

$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{n!(n+1)!4^{(2n+3)}}$$

$$\phi=2\cos (\pi/5)=2\sum_{k=0}^\infty \frac{((-1)^k (\pi/5)^{2 k}}{(2k)!}$$

$$\phi=\frac{1}{2}+\frac{\sqrt{5}}{2}=\frac{1}{2}+\sum_{n=0}^\infty 4^{-n}\binom{1/2}{n}$$

Not exactly a series, but might also be of interest:

$$1-\frac{1}{\phi}=\frac{1}{\phi^2}=\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\frac{1}{5} \left(1+\dots \right)^2 \right)^2 \right)^2 \right)^2$$

$$\frac{1}{\phi^4}=\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\frac{1}{5} \left(1-\dots \right)^2 \right)^2 \right)^2 \right)^2$$

$$\frac{1}{\phi^4}=\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\frac{1}{9} \left(1+\dots \right)^2 \right)^2 \right)^2 \right)^2$$

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:

$$\prod_{n=1}^{\infty}\left[(-1)^{n+1}\phi F_n+(-1)^nF_{n+1}\right]^{n^{-(s+1)}}=\phi^{-\zeta(s)}$$

• This is Brilliant!!! Commented May 4, 2016 at 9:28

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 $$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(\phi+y^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{\phi^2}$$

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

$$\fbox{\Large\int_0^\infty{x^{-7/10}\over1+x}\mathrm dx={2\pi\over\phi}}$$

-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 $\text{"something"}$