Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\phi$?


$^*$ 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.

share|cite|improve this question
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

27 Answers 27

Potentially interesting:


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:



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:

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

share|cite|improve this answer
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 $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi} $$

share|cite|improve this answer
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:

artist view of the identity

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.

share|cite|improve this answer
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

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

share|cite|improve this answer
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 $\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \phi$.

share|cite|improve this answer
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 $\varphi = 2\cos(\frac{\pi}{5})$:

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

share|cite|improve this answer
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

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

share|cite|improve this answer
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.

share|cite|improve this answer

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


share|cite|improve this answer
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

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

share|cite|improve this answer
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

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


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.

share|cite|improve this answer
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:… – 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.

$$\int_0^{\pi/2} \ln(1+4\sin^2 x)\text{ d}x=\pi\log\left(\varphi\right)$$


$$\int_0^{\pi/2} \ln(1+4\sin^4 x)\text{ d}x=\pi\log \frac{\varphi+\sqrt{\varphi}}{2}$$

Again, not mine. But they definitely deserve to be here

share|cite|improve this answer
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

$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large \int_0^\infty \ln \left( \frac{x^2+2kx\cos \sqrt{a^2+\phi}+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}=\phi}}$$ I hope you find this integral interesting.

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

share|cite|improve this answer
Click the box for the proof – Venus Jun 1 at 8:41

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:

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

share|cite|improve this answer
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:

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

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

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

This one is a bit messy.

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

share|cite|improve this answer

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

share|cite|improve this answer
Awesome one!!!! – Beta May 4 at 9:27

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

share|cite|improve this answer

Consider the sequence


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



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 $a_n$.

share|cite|improve this answer
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:

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

share|cite|improve this answer
This is Brilliant!!! – Beta May 4 at 9:28

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

share|cite|improve this answer

-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 $\text{"something"}$

share|cite|improve this answer

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


share|cite|improve this answer

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

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

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

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

share|cite|improve this answer

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

share|cite|improve this answer

protected by Ron Gordon Feb 15 at 20:40

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.