I found the following identity. While trying to prove it, I found some things that I don’t quite understand:

$$\frac{\pi}{4}=\sqrt{5} \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}$$

(where $\phi=\frac{\sqrt{5}+1}{2}$).

## What I tried

I first considered the series: $$F(x)=\sum_{n=0}^{\infty}F_{2n+1}x^{n}=\frac{1-x}{x^2-3x+1}$$ (when it converges).

Then I replaced $x$ with $x^2$ and tried integrating to get something like:

$$A(x)=\sum_{n=0}^{\infty}F_{2n+1} \frac{x^{2n+1}}{2n+1}=\int \frac{1-x^2}{x^4-3x^2+1}$$

This is where one question arises:

- Is this integration a valid thing to do? The sum on the left has a value, but the integral on the right has some constant added to it. In that case, how should I choose the constant?

Now, making $x=\frac{1}{\phi^2}$ would make the $\phi^{4n+2}$ term appear, but we still need to put a $(-1)^n$ in there, so I thought of putting an $i$ there (because $i^{2n+1}=i \cdot (-1)^n$) this way:

$x=\frac{i}{\phi^2}$, and then $$A(x)=i \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}.$$ So now I need to prove that the integral for this $x$ is exactly $\frac{i \pi}{4\sqrt{5}}$, but the problem I have is that the integral has logarithms and I don’t know how to find logarithms of complex numbers like $\log(5+2i)$. (I found on Wikipedia the Taylor series for logarithms, but I can’t see how this makes the problem simpler.)

## More questions

Does it make sense to plug this complex value into the power series and the integral? If so, then how one would evaluate the integral for this specific $x=\frac{i}{\phi^2}$?

Is there any other path to prove this intriguing identity?