Showing $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \log(\cos(\phi))\cos(\phi) \ d\phi = \log(4) - 2 $

Question

This is a minor detail of a proof in 'Chaotic Billiards' by Chernov and Markarian which I foolishly decided to verify.

It's page 44 of the book, during the proof that lyapunov exponents exist almost everywhere in the collision space (Theroem 3.6) in case anyone is familiar. Though the integral problem stated is independent of any knowledge of this field.

I've tried integration by parts but I end up with:

$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin(\phi)\tan(\phi) \,d\phi$

which is non-convergent on the given interval. All help is appreciated!

quick edit: log is the natural logarithm

Answer 1

You can replace $\cos(x)$ inside the log with $\sqrt{1-\sin(x)^2}$, then use the properties of the logarithm to expand it into:

$$\dfrac{1}{2}\int (\log(1-\sin(x)) + \log(1+\sin(x)))\cos(x) dx$$

Now you can integrate the individual summands using $u$-substitution.

Answer 2

Here is one alternate method. Since $$ \log(\cos(\theta))=\frac12\left[\log\left(1+e^{2i\theta}\right)+\log\left(1+e^{-2i\theta}\right)-\log(4)\right] $$ we get $$ \begin{align} &\int_{-\pi/2}^{\pi/2}\left[-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2k\theta)}{k}\right]\cos(\theta)\,\mathrm{d}\theta\\ &=-2\log(2)-\sum_{k=1}^\infty(-1)^k\int_{-\pi/2}^{\pi/2}\frac{\cos((2k+1)\theta)+\cos((2k-1)\theta)}{2k}\mathrm{d}\theta\\ &=-2\log(2)-\sum_{k=1}^\infty\left[\frac1{k(2k+1)}-\frac1{k(2k-1)}\right]\\ &=-2\log(2)-\sum_{k=1}^\infty\left[\color{#C00000}{\frac2{2k}-\frac2{2k-1}}+\color{#00A000}{\frac2{2k}-\frac2{2k+1}}\right]\\[6pt] &=-2\log(2)-[\color{#C00000}{-2\log(2)}+\color{#00A000}{2-2\log(2)}]\\[16pt] &=2\log(2)-2 \end{align} $$

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Integration by parts works as well. but it requires more than one application. $$ \begin{align} &\int\log(\cos(\theta))\cos(\theta)\,\mathrm{d}\theta\\ &=\int\log(\cos(\theta))\,\mathrm{d}\sin(\theta)\\ &=\sin(\theta)\log(\cos(\theta))+\int\sin(\theta)\tan(\theta)\,\mathrm{d}\theta\\ &=\sin(\theta)\log(\cos(\theta))-\int\tan(\theta)\,\mathrm{d}\cos(\theta)\\ &=\sin(\theta)\log(\cos(\theta))-\sin(\theta)+\int\cos(\theta)\sec^2(\theta)\,\mathrm{d}\theta\\ &=\sin(\theta)\log(\cos(\theta))-\sin(\theta)+\int\frac1{1-\sin^2(\theta)}\,\mathrm{d}\sin(\theta)\\[3pt] &=\sin(\theta)\log(\cos(\theta))-\sin(\theta)+\log(\sec(\theta)+\tan(\theta))+C \end{align} $$ Since $\log(\sec(\theta)+\tan(\theta))=-\log(\sec(\theta)-\tan(\theta))$, we can evaluate the antiderivative at $+\pi/2$ and $-\pi/2$: $$ \begin{align} \text{at }+\pi/2:&-1+\log(2)+C\\ \text{at }-\pi/2:&+1-\log(2)+C \end{align} $$ Thus, we get that $$ \int_{-\pi/2}^{\pi/2}\log(\cos(\theta))\cos(\theta)\,\mathrm{d}\theta=2\log(2)-2 $$

Answer 3

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

\begin{align}&\color{#00f}{\large% \int_{-\pi/2}^{\pi/2}\ln\pars{\cos\pars{\phi}}\cos\pars{\phi}\,\dd\phi} =2\int_{0}^{\pi/2}\ln\pars{\cos\pars{\phi}}\cos\pars{\phi}\,\dd\phi \\[3mm]&=2\int_{1}^{0}\ln\pars{t}t\,\pars{-\,{\dd t \over \root{1 - t^{2}}}} =2\lim_{\mu \to 1}\totald{}{\mu}\int_{0}^{1}t^{\mu}\pars{1 - t^{2}}^{-1/2}\,\dd t \\[3mm]&=2\lim_{\mu \to 1}\totald{}{\mu} \int_{0}^{1}t^{\mu/2}\pars{1 - t}^{-1/2}\,\half\,t^{-1/2}\dd t =\lim_{\mu \to 1}\totald{}{\mu} \int_{0}^{1}t^{\pars{\mu - 1}/2}\pars{1 - t}^{-1/2}\,\dd t \\[3mm]&=\lim_{\mu \to 1}\totald{{\rm B}\pars{\bracks{\mu + 1}/2,1/2}}{\mu} =\lim_{\mu \to 1}\totald{}{\mu} \bracks{\Gamma\pars{\bracks{\mu + 1}/2}\Gamma\pars{1/2} \over \Gamma\pars{\mu/2 + 1}} \\[3mm]&=\Gamma\pars{\half}\braces{% {\Gamma\pars{1} \over \Gamma\pars{3/2}}\, \half\bracks{\Psi\pars{1} - \Psi\pars{3 \over 2}}} =\Gamma\pars{\half}\braces{{1 \over \Gamma\pars{1/2}/2}\, \half\bracks{\ln\pars{4} - 2}} \\[3mm]&= \color{#00f}{\large\ln\pars{4} - 2} \end{align}

$\ds{{\rm B}\pars{x,y}}$ is the Beta Function. $\ds{\Gamma\pars{z}}$ is the Gamma Function. $\ds{\Psi\pars{z}}$ is the Digamma Function.

We used well known properties of those functions.