# 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). – Clement C. Feb 14 '16 at 3:15
• Related question introducing an infinite product for GR. And this question – Yuriy S Feb 14 '16 at 3:32
• Also this. Somewhat famous locally :-) – Jyrki Lahtonen Feb 14 '16 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. is a poor mathematician Feb 15 '16 at 14:31
• Hey guys could we get done proofs of these integrals please? – Faraz Masroor Feb 16 '16 at 12:44

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

• Could those people vote to delete please explain the reason? – Kemono Chen Oct 27 '18 at 13:02

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! – Von Neumann Feb 10 at 21:37

$$\int_{0}^{\phi}(1-x+x^2)^{1/\phi}(1-\phi^2x+\phi^3x^2)\mathrm dx=2^{\phi}\cdot\phi$$

A bit over-crowed in term of $\phi$

$$\int_0^1 \frac{2}{\left(1+\phi x^2\right) \sqrt{1-x^2}} \, dx=\frac{\pi }{\phi }$$

This is my attempt to find another integral representation for Golden ratio using some special function as shown in the above image !!!!!!!

• I already saw it in a past question of yours. It's really cool, can you prove it? – Von Neumann Mar 22 '18 at 6:42
• Can we see all the proof please ? – Abr001am May 8 '18 at 15:28

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

$$\int_{0}^{1}{1-x^{\phi}\over 1-x}+{x(1-\phi x^{1\over \phi}+{1\over \phi}x^{\phi})\over (1-x)^2}\mathrm dx=\phi$$

## protected by Ron GordonFeb 15 '16 at 20:40

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