Fibonacci infinite sum resulting in $\pi$ 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?

 A: By the explicit formula for Fibonacci numbers it follows that:
$$\color{red}{S}=\color{blue}{\sqrt{5}}\sum_{n\geq 0}\frac{(-1)^n \color{blue}{F_{2n+1}}}{(2n+1)\,\color{blue}{\varphi^{4n+2}}}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left(\color{blue}{\frac{1}{\varphi^{2n+1}}+\frac{1}{\varphi^{6n+3}}}\right),$$
hence by the arctangent Taylor series and the (arc)tangent sum formulas:
$$ \color{red}{S} = \arctan\frac{1}{\varphi}+\arctan\frac{1}{\varphi^3}=\arctan\frac{\frac{1}{\varphi}+\frac{1}{\varphi^3}}{1-\frac{1}{\varphi^4}}=\arctan 1=\color{red}{\frac{\pi}{4}}$$
as wanted.
A: Start with
$$
\begin{align}
\sum_{k=0}^\infty F_{2k+1}x^{2k}
&=\frac{1-x^2}{1-3x^2+x^4}\\
&=\frac{1-x^2}{(x-\phi)(x+\phi)(x+\frac1\phi)(x-\frac1\phi)}\\
&=\frac1{2\sqrt5}\frac1{x+\phi}-\frac1{2\sqrt5}\frac1{x-\phi}+\frac1{2\sqrt5}\frac1{x+\frac1\phi}-\frac1{2\sqrt5}\frac1{x-\frac1\phi}\tag{1}
\end{align}
$$
Integration yields
$$
\begin{align}
\sum_{k=0}^\infty\frac{F_{2k+1}}{2k+1}x^{2k+1}
&=\frac1{2\sqrt5}\log\left(\frac{(x+\phi)(x+\frac1\phi)}{(x-\phi)(x-\frac1\phi)}\right)\\
&=\frac1{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right)\tag{2}
\end{align}
$$
Substituting $x\mapsto ix$ then multiplying by $-i\sqrt5$ gives
$$
\begin{align}
\sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}x^{2k+1}
&=\frac1{2i}\log\left(\frac{1-x^2+i\sqrt5x}{1-x^2-i\sqrt5x}\right)\\
&=\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)+i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\
&-\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)-i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\
&=\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\tag{3}
\end{align}
$$
since $\log(a+ib)=\tfrac12\log(a^2+b^2)+i\arctan\left(\tfrac ba\right)$ for $a\gt0$; that is $|x|\lt1$.
Evaluating $(3)$ at $x=\frac1{\phi^2}$ yields
$$
\begin{align}
\sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}\frac1{\phi^{4k+2}}
&=\arctan\left(\frac{\sqrt5}{\phi^2-\frac1{\phi^2}}\right)\\
&=\arctan(1)\\[9pt]
&=\frac\pi4\tag{4}
\end{align}
$$
A: 
1.Is this integration a valid thing to do? because the sum on the left has a value but the integral on the right has some constant added to it .Then how to choose the constant ?

Answer: Yes, integration is a perfectly valid step in a solution to problems like this, but not indefinite integration. The integration must be definite, otherwise the equality is between a function on the LHS:
$$
\sum_{n=0}^\infty{}F_{2n+1}\frac{x^{2n+1}}{2n+1}
$$
and a set of functions on the RHS $$
\left\{c + \int\frac{1-x^2}{x^4-3x^2+1}\text{d}x,\ c\in\Bbb{R}\right\}
$$ which doesn't make any sense. 
So, we have to choose appropriate limits. To choose these limits, think about what you did to the LHS to go from $\sum{F_{2n+1}x^{2n}}$ to $\sum{F_{2n+1}\frac{x^{2n+1}}{2n+1}}$. We integrated the former sum from $0$ to $x$:
$$
\int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t = \sum_{n=0}^\infty{F_{2n+1}\frac{x^{2n+1}}{2n+1}}
$$
but since the sum in the integrand is equal to $\frac{1-x^2}{x^4-3x^2+1}$, then we also have
$$
\begin{align}
\int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t & = \int_0^x\frac{1-t^2}{t^4-3t^2+1}\text{d}t \\
& = \frac{1}{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right) \equiv \text{A}(x)
\end{align}
$$
From here, robjohn's answer explains the rest of the proof.
