Vtgyjfy

Potentially interesting:

$$logvarphi=int_0^{1/2}frac{dx}{sqrt{x^2+1}}$$

Perhaps also worthy of consideration:

$$arctan frac{1}{varphi}=frac{int_0^2frac{1}{1+x^2}, dx}{int_0^2 dx}=frac{int_{-2}^2frac{1}{1+x^2}, dx}{int_{-2}^2 dx}$$

A development of the first integral:

$$logvarphi=frac{1}{2n-1}int_0^{frac{F_{2n}+F_{2n-2}}{2}}frac{dx}{sqrt{x^2+1}}$$

$$logvarphi=frac{1}{2n}int_1^{frac{F_{2n+1}+F_{2n-1}}{2}}frac{dx}{sqrt{x^2-1}}$$

which stem from the relationship $(x-varphi^m)(x-barvarphi^m)=x^2-(F_{m-1}+F_{m+1})x+(-1)^m$, where $barvarphi=frac{-1}{varphi}=1-varphi$ and $F_k$ is the $k$th Fibonacci number. I particularly enjoy:

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

In this answer, it is shown that
$$
int_0^inftyfrac{sqrt{x}}{x^2+2x+5}mathrm{d}x=fracpi{2sqrtphi}
$$

[EDIT:20190112 Another connection between $pi$ and $phi$ at the bottom] An identity derived from the Rogers-Ramanujan continued fraction ($R(q)$, not defined here) exhibits a $phi$ factor:

$$ frac{1}{(sqrt{phisqrt{5}})e^{2pi/5}} = 1+frac{e^{-2pi}}{1+frac{e^{-4pi}}{1+frac{e^{-6pi}}{1+frac{e^{-8pi}}{1+frac{e^{-10pi}}{1+frac{e^{-12pi}}{cdots}}}}}}$$

and one can then obtain a formula like:
$$ ln left( sqrt{4phi+3}-phi^2right) = -frac{1}{5}int_{e^{-2pi}}^1 frac{(1-t)^5(1-t^2)^5(1-t^3)^5 dots}{(1-t^5)(1-t^{10})(1-t^{15}) dots}frac{dt}{t}$$
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.

[EDIT] In M. D. Hirschhorn, A connection between $pi$ and $phi$, Fibonacci Quarterly, 2015, another asymptotic relation is:

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

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

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

Based on the fact that $varphi = frac{1+sqrt{5}}{2}$:

$$varphi = int_4^5 frac32+frac1{4sqrt{x}} mathrm{d}x$$

Based on the fact that $varphi = 2cos(frac{pi}{5})$:

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

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

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

All the following is based on the simple fact that:

$$phi=2 cos left( frac{pi}{5} right)=2 sin left( frac{3pi}{10} right)$$

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.

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

$$int_{-infty}^0sqrt{rho^2+dotrho^2}dtheta=int_{-infty}^0sqrt{1+2^2}e^{2theta}dtheta=phi-frac12.$$

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:

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

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

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.

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.

$$int_0^{pi/2} ln(1+4sin^2 x)text{ d}x=pilogleft(varphiright)$$

and

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

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

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

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 $a_n$.

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

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

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

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

$$
int_0^infty frac{1}{1+x^{10}}dx=frac{phipi}{5}
$$

-I remember really liking this one:

$$int_0^1 int_0^1 frac{text{dx dy}}{varphi^6-x^2y^2}=frac{pi^2-18log^2varphi}{24varphi^3}$$

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 $text{"something"}$

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

This one is a bit messy.

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

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

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:

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

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

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

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

Here is a collection of the series with reciprocal binomial coefficients.

$$sum_{n=0}^infty (-1)^n left( begin{matrix} 2n \ n end{matrix} right)^{-1}=frac{4}{5} left(1-frac{sqrt{5}}{5} ln phi right)$$

$$sum_{n=1}^infty frac{(-1)^n}{n} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=-frac{2sqrt{5}}{5} ln phi$$

$$sum_{n=1}^infty frac{(-1)^n}{n^2} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=-2 ln^2 phi$$

$$sum_{n=0}^infty frac{(-1)^n}{2n+1} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=frac{4sqrt{5}}{5} ln phi$$

$$sum_{n=0}^infty frac{(-1)^n}{n+1} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=frac{8sqrt{5}}{5} ln phi-4 ln^2 phi$$

$$sum_{n=2}^infty frac{(-1)^n}{n-1} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=frac{3sqrt{5}}{5} ln phi-frac{1}{2}$$

$$sum_{n=2}^infty frac{(-1)^n}{(n-1)^2} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=1-sqrt{5} ln phi+ ln^2 phi$$

$$sum_{n=2}^infty frac{(-1)^n}{n^2(n^2-1)} left( begin{matrix} 2n \ n end{matrix} right)^{-1}=4ln^2 phi-frac{sqrt{5}}{2} ln phi-frac{3}{8}$$

A one with $pi$:

$$sum_{n=0}^infty left( begin{matrix} 4n \ 2n end{matrix} right)^{-1}=frac{16}{15}+frac{sqrt{3}}{27} pi-frac{2sqrt{5}}{25} ln phi $$

Source here

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

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

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

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

Here is another one

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

We can prove the inequalities

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

with representations

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

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