$\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio

Question

We already know that, for any real number $t$ such that $t\geq-1$,

$$ \int_{0}^{\pi/2} \ln \left(1+t \sin^2 x\right) \mathrm{d}x = \pi \ln \left( \frac{1+\sqrt{1+t}}{2} \right). $$

Prove that

$$ \int_{0}^{\pi/2} \ln \left(1+4\sin^4 x\right) \mathrm{d}x = \pi \ln \left( \frac{\varphi+\sqrt{\varphi}}{2} \right) $$ where $\displaystyle \varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.

Answer 1

Hint: $\quad1+t^2\sin^4x~=~(1-ti\sin^2x)\cdot(1+ti\sin^2x),\qquad\log(ab)=\log a+\log b,\quad$ and

$\Re(\pm~ti)>-1$.

Answer 2

Let's put it with real numbers.

We have

$$ \int_{0}^{\pi/2} \ln \left(1+t\sin^4 x\right) \mathrm{d}x = \pi \ln \left( \frac{1}{4} \sqrt{ 1 + \sqrt{1+t} } \left( \sqrt{1 + \sqrt{1+t} } + \sqrt{2} \right) \right), \quad t \geq -1. $$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/2}\ln\pars{1 + 4\sin^{4}\pars{x}}\,\dd x =\pi\ln\pars{\varphi + \root{\varphi} \over 2}:\ {\large ?}}$

\begin{align}& \partiald{}{\mu}\int_{0}^{\pi/2}\ln\pars{1 + \mu\sin^{4}\pars{x}}\,\dd x =\int_{0}^{\pi/2}{\sin^{4}\pars{x} \over 1 + \mu\sin^{4}\pars{x}}\,\dd x \\[3mm]&={\pi \over 2\mu} -{1 \over \mu}\color{#c00000}{\int_{0}^{\pi/2} {\dd x \over 1 + \mu\sin^{4}\pars{x}}} \end{align}

\begin{align}&\color{#c00000}{\int_{0}^{\pi/2}% {\dd x \over 1 + \mu\sin^{4}\pars{x}}} =\int_{0}^{\pi/2}{\csc^{4}\pars{x} \over \csc^{4}\pars{x} + \mu}\,\dd x =\ \overbrace{\int_{0}^{\pi/2}% {\cot^{2}\pars{x} + 1 \over \bracks{\cot^{2}\pars{x} + 1}^{2}+ \mu} \,\csc^{2}\pars{x}\,\dd x}^{\ds{\cot\pars{x} \equiv t}} \\[3mm]&=-\int_{\infty}^{0}{t^{2} + 1 \over \pars{t^{2} + 1}^{2} + \mu}\,\dd t =\Re\int_{0}^{\infty}{\dd t \over t^{2} + 1 + \root{\mu}\ic} \\[3mm]&=\Re\bracks{{1 \over \root{1 + \root{\mu}{\ic}}} \int_{0}^{\infty/\root{1 + \root{\mu}{\ic}}}{\dd t \over t^{2} + 1}} ={\pi \over 2}\,\Re\pars{{1 \over \root{1 + \root{\mu}{\ic}}}} \end{align}

\begin{align}&\color{#66f}{% \large\int_{0}^{\pi/2}\ln\pars{1 + 4\sin^{4}\pars{x}}\,\dd x} ={\pi \over 2}\,\Re\int_{0}^{4}\overbrace{% {1 \over \mu}\pars{1 - {1 \over \root{1 + \root{\mu}\ic}}}\,\dd\mu} ^{\ds{t \equiv \root{1 + \root{\mu}\ic}}} \\[3mm]&=2\pi\,\Re\int_{1}^{\root{1 + 2\,\ic}}{\dd t \over 1 + t} =\color{#66f}{\large 2\pi\,\Re\ln\pars{1 + \root{1 + 2\ic} \over 2}} \approx 1.1565078476153109133 \end{align}

It agrees with the OP proposed answer $\color{#000}{\large\quad\ds{\pi\,\ln\pars{\varphi + \root{\varphi} \over 2}}}$.