Is there any integral for the Golden Ratio?

Solution 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:

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

Solution 2:

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