# Is there any integral for the Golden Ratio?

This is a curiosity. I was wondering about math important/famous constants, like $e$, $\pi$, $\gamma$ and obviously $\phi$.

The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:

$$\pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$

$$e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$

$$\gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$

My question is: is there some interesting integral $^*$ (or also some series) whose result is simply $\phi$?

Note

$^*$ 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.

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You can skim this page, on WolframAlpha; e.g. Eq (12) and (13). – Clement C. Feb 14 at 3:15
Related question introducing an infinite product for GR. And this question – Yuriy S Feb 14 at 3:32
Also this. Somewhat famous locally :-) – Jyrki Lahtonen Feb 14 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. – J. M. Feb 15 at 14:31
Hey guys could we get done proofs of these integrals please? – Faraz Masroor Feb 16 at 12:44

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:

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

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Wow. Did you come up with this by yourself ? – user230452 Feb 14 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 Feb 14 at 4:28
What about $$\int_0^{1/2}\left(\frac{x}{\sqrt{x^2+1}}+3\right)\,dx$$ – Yves Daoust Feb 14 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 $k\geq2$. – David Richerby Feb 14 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 Feb 16 at 14:00

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

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Awesome!! A strict link between $\pi$ and $\phi$, I love those things. Thank you! – Beta Feb 14 at 14:35
Brilliant!! Absolutely amazing – Albas Feb 14 at 15:04
wow! this is incredible – Andres Mejia Feb 14 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". – MichaelS Feb 14 at 22:36
So very nice ! Somehow you perhaps can rope in $e$ too. – Narasimham Feb 15 at 15:18

An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $\phi$ factor:

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.

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The genius of Ramanujan will always remain a mystery.. what a genius. – Beta Feb 15 at 14:33
And I believe it is a good thing that this remains a mystery. – Laurent Duval Feb 15 at 14:58
mind... blown... – MichaelChirico Feb 18 at 5:03

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

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There is a sign error in the log term – Laplacian Fourier Feb 14 at 16:37
@LaplacianFourier: Thanks. – Ron Gordon Feb 14 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.. – Faraz Masroor Feb 14 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! – Ron Gordon Feb 14 at 21:51
...might as well include a link: MSE 562964 – Benjamin Dickman Feb 18 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 $\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \phi$.

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Was this the first definition of golden ratio or did it have a definition before that ? – user230452 Feb 14 at 4:27
@user230452 $\phi = \frac { 1+ \sqrt 5}2$ – Ant Feb 14 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 ? – user230452 Feb 14 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. – Wojowu Feb 14 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. – David Richerby Feb 14 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 $\varphi = 2\cos(\frac{\pi}{5})$:

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

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

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

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Great! Another integral that relates two constants! Thank you! – Beta Feb 14 at 19:37
@KimPeek, there is an infinite number of integrals of this kind – Yuriy S Feb 14 at 19:42
@YuriyS The more I see, the happier I am :D – Beta Feb 14 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) – Laurent Duval Feb 15 at 7:14

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.

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

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

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

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

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Astounding beauty – Beta Apr 24 at 11:04
Another integral involving $\phi$ that might be surprising at first sight :) $$\int_0^\infty\frac1{1+x^2}\frac1{1+x^\phi}dx=\frac\pi4.$$ – Vladimir Reshetnikov May 27 at 19:27

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

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.

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

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

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Beautiful! Thank you for having posted them here. The first one is so beautiful!! – Beta Apr 2 at 13:42

For $a=\sqrt{\pi^2-\phi}$ and $k>0$, we have

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

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Click the box for the proof – Venus Jun 1 at 8:41

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

There is an infinitely nested radical in the denominator.

A finite one is also possible:

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

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

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

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

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

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

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

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Here is another one $$\int_0^\infty \frac{1}{5^{\frac{x}{4}}+5^{\frac{1}{2}}-5^0}dx=\phi$$

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This one is a bit messy.

$$\int_0^\infty \frac{1}{(\sqrt5^x)^{2^{-(\sqrt5-1)}}+\sqrt5-1}dx=2^{\phi^{-3}}\cdot\phi$$

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

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Awesome one!!!! – Beta May 4 at 9:27

$$\int_0^\infty \frac{1}{1+x^{\frac{10}{3}}}dx=\frac{3\pi}{5\phi}$$

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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 $a_n$.

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

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:

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

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This is Brilliant!!! – Beta May 4 at 9:28

$$\int_0^1 \frac{200\sqrt5(1-x^2)-300(1-x)^2}{ \left[5\sqrt5(1+x)^2-15(1-x^2)+2\sqrt5(1-x)^2 \right]^2}dx=(2\phi+1)(\phi+2)$$

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-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 $\text{"something"}$

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Notice that $\frac{2}{1+\sqrt5}=\frac{1}{\phi}$

$$\int_0^1\frac{2}{(1+\sqrt5x)^2}dx=\frac{1}{\phi}$$

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

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

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

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

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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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