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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.

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    $\begingroup$ You can skim this page, on WolframAlpha; e.g. Eq (12) and (13). $\endgroup$ – Clement C. Feb 14 '16 at 3:15
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    $\begingroup$ Related question introducing an infinite product for GR. And this question $\endgroup$ – Yuriy S Feb 14 '16 at 3:32
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    $\begingroup$ Also this. Somewhat famous locally :-) $\endgroup$ – Jyrki Lahtonen Feb 14 '16 at 9:45
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    $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Feb 15 '16 at 14:31
  • $\begingroup$ Hey guys could we get done proofs of these integrals please? $\endgroup$ – Faraz Masroor Feb 16 '16 at 12:44

37 Answers 37

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

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  • $\begingroup$ Could those people vote to delete please explain the reason? $\endgroup$ – Kemono Chen Oct 27 '18 at 13:02
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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 $\pi=3.14...$

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  • $\begingroup$ This is awesome! $\endgroup$ – Von Neumann Feb 10 at 21:37
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$$\int_{0}^{\phi}(1-x+x^2)^{1/\phi}(1-\phi^2x+\phi^3x^2)\mathrm dx=2^{\phi}\cdot\phi$$

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

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

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enter image description here

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

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    $\begingroup$ I already saw it in a past question of yours. It's really cool, can you prove it? $\endgroup$ – Von Neumann Mar 22 '18 at 6:42
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    $\begingroup$ Can we see all the proof please ? $\endgroup$ – Abr001am May 8 '18 at 15:28
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$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(1+x+x^2)^2+x^2}=\pi\cdot \sqrt{\frac{\phi}{5}}$$

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

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protected by Ron Gordon Feb 15 '16 at 20:40

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