Is there any integral for the Golden Ratio?

Question

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.

Answer 1

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:

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

Answer 2

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

Answer 3

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:

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

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

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

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

Answer 12

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:

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

Answer 13

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

Answer 14

Five:Pi:Phi

Coincidence closed form

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

Answer 15

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

---

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.

Answer 16

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

Answer 17

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 $\pi=3.14...$

Answer 18

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

Answer 19

Here is another one

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

Answer 20

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

Answer 21

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 $a_n$.

Answer 22

We can prove the inequalities

$$\frac{3}{2}<\frac{8}{5}<\phi<\frac{13}{8}<\frac{5}{3}$$

with representations

$$\begin{align}\phi&=\frac{3}{2}+\frac{1}{4}\int_0^1 \frac{dx}{\sqrt{4+x}}\\ \\ \phi&=\frac{8}{5}+\frac{1}{5}\int_0^1 \frac{dx}{\sqrt{121+4x}}\\ \\ \phi&=\frac{13}{8}-\frac{1}{16}\int_0^1 \frac{dx}{\sqrt{80+x}}\\ \\ \phi&=\frac{5}{3}-\frac{1}{3}\int_0^1 \frac{dx}{\sqrt{45+4x}}\\ \end{align}$$

Answer 23

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

Answer 24

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

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

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

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

Answer 25

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

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

---

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

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

Answer 26

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:

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

Answer 27

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

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

Answer 29

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

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

Answer 30

-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 $\text{"something"}$