Evaluate $\int_{0}^{\frac{\pi}{4}}\ln(\cos t)dt$

Question

$$\int_{0}^{\frac{\pi}{4}}\ln(\cos t)dt=-\frac{{\pi}}{4}\ln2+\frac{1}{2}K$$

I ran across this integral while investigating the Catalan constant. I am wondering how it is evaluated. I know of this famous integral when the limits of integration are $0$ and $\frac{\pi}{2}$, but when the limits are changed to $0$ and $\frac{\pi}{4}$, it becomes more complicated.

I tried using $$\cos(t)=\frac{e^{it}+e^{-it}}{2},$$ then rewriting it as:
$$\int_{0}^{\frac{\pi}{4}}\ln\left(\frac{e^{it}+e^{-it}}{2}\right)=\int_{0}^{\frac{\pi}{4}}\ln(e^{it}+e^{-it})dt-\int_{0}^{\frac{\pi}{4}}\ln(2)dt.$$

But, this is where I get stuck.

Maybe factor out an $e^{it}$ and get $$\int_{0}^{\frac{\pi}{4}}\ln(e^{it}(1+e^{-2it}))dt=\int_{0}^{\frac{\pi}{4}}\ln(e^{it})dt+\int_{0}^{\frac{\pi}{4}}\ln(1+e^{-2it})dt$$

I thought maybe the Taylor series for $\ln(1+x)$ may come in handy in some manner. And it is

$$\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k}}{k}$$

which gives

$$\int_{0}^{\frac{\pi}{4}}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}e^{-2kit}}{k}$$

Just some thoughts. I doubt if I am on to anything. I used a technique similar to this when solving $\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$.

But, how in the world would the Catalan constant come into the solution:

$$K=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx 0.916$$

Your learned input is appreciated. Thanks!

Answer 1

I dont know what a Catalan number is but the real part of the integral gets you the expression you are seeking. You have $$\int_{0}^{\frac{\pi}{4}}\ln(e^{it}+e^{-it})dt= i(2n\frac{\pi^2}{4}+\frac{\pi^2}{16}) + \int_{0}^{\frac{\pi}{4}} \left[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}e^{-2kit}}{k}\right]dt$$

Here $n$ just appears because of the multivalued $\ln (e^{it})$ term but you can ignore it as they are imaginary. The expression you got correctly was

$$\int_{0}^{\frac{\pi}{4}}\left(\sum_{k=1}^{\infty}\frac{(-1)^{k+1}e^{-2kit}}{k}\right)dt$$

$$=\sum_{k=1}^{\infty}\left(\int_{0}^{\frac{\pi}{4}}\frac{(-1)^{k+1}e^{-2kit}}{k}dt\right)$$

$$=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(e^{-i\frac{k\pi}{2}}-1)}{(-2ki)k}= \sum_{k=1}^{\infty}\frac{(-1)^{k+1}(e^{-i\frac{k\pi}{2}})}{(-2ki)k}-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(-2ki)k}$$

The second summation in this expression is gonna remain imaginary forever, so leave it out. In the first term observe that the summand is going to be real only if $k=2n+1$ as n goes from $0$ to $\infty$.

Substituting $k=2n+1$, and $e^{i(2n+1)\frac{\pi}{2}}=(-1)^n i$ you get the real part of the entire thing as

$$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2(2n+1)^{2}} =\frac{K}{2}$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t & = -\,{\pi\ln\pars{2} \over 4} + {K \over 2}:\ {\large ?} \\[2mm] \mbox{where}\ K & \equiv\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} \approx 0.9160 \end{align} is the Catalan Constant.

\begin{align} &\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t} =-\,{\pi\ln\pars{2} \over 4} + \int_{0}^{\pi/4}\ln\pars{2\cos\pars{t}}\,\dd t \\[5mm]=& \ -\,{\pi\ln\pars{2} \over 4} + \half\int_{0}^{\pi/4}\ln\pars{\cot\pars{t}} \,\dd t \\[2mm] & + \int_{0}^{\pi/4}\ln\pars{2\cos\pars{t} \over \cot^{1/2}\pars{t}}\,\dd t \end{align} Since ( see this link ) $\ds{K = \int_{0}^{\pi/4}\ln\pars{\cot\pars{t}}\,\dd t}$: \begin{align} & \color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t} \\[5mm] = & \ -\,{\pi\ln\pars{2} \over 4} + {K \over 2} + \half \color{#00f}{\int_{0}^{\pi/4}\ln\pars{4\cos^{2}\pars{t} \over \cot\pars{t}}\,\dd t} \end{align} The problem is reduced to show that the "$\color{#00f}{\mbox{blue integral}}$" vanishes out: \begin{align} &\color{#00f}{\int_{0}^{\pi/4}\ln\pars{4\cos^{2}\pars{t} \over \cot\pars{t}}\,\dd t} \\[5mm] = & \ \int_{0}^{\pi/4}\ln\pars{4\sin\pars{t}\cos\pars{t}}\,\dd t \\[5mm] = & \ \int_{0}^{\pi/4}\ln\pars{2\sin\pars{2t}}\,\dd t \\[5mm] = & \ \half\int_{0}^{\pi/2}\ln\pars{2\sin\pars{t}}\,\dd t \\[5mm] = & \ {1 \over 4}\,\pi\ln\pars{2} + \half\,\lim_{\mu \to 0}\partiald{}{\mu} \int_{0}^{1}t^{\mu}\pars{1 - t^{2}}^{-1/2}\,\dd t \\[5mm] = & \ {1 \over 4}\,\pi\ln\pars{2} + {1 \over 4}\,\lim_{\mu \to 0}\partiald{}{\mu} \int_{0}^{1}t^{\pars{\mu - 1}/2}\pars{1 - t}^{-1/2}\,\dd t \\[5mm]= & \ {1 \over 4}\,\pi\ln\pars{2} + {1 \over 4}\,\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \Gamma\pars{\mu/2 + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{\mu/2 + 1}} \\[5mm] = & \ {1 \over 4}\,\pi\ln\pars{2} \\ & + {1 \over 8}\,{\Gamma\pars{1/2} \over \Gamma\pars{1}}\ \bracks{\overbrace{\Psi\pars{\half} - \Psi\pars{1}} ^{\ds{-2\ln\pars{2}}}}\, \overbrace{\Gamma\pars{\half}}^{\ds{\root{\pi}}} \\[5mm] = & \ \color{#00f}{\large 0} \quad\mbox{since}\quad\Gamma\pars{1} = 1 \end{align} $\ds{\Gamma\pars{z}}$ and $\ds{\Psi\pars{z}}$ are the Gamma and Digamma Functions, respectively. $\pars{1}$ and $\pars{2}$ lead to: $$ \color{#00f}{\large\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t =-\,{\pi\ln\pars{2} \over 4} + {K \over 2}} $$

Answer 3

With $\int_{0}^{\frac{\pi}{4}}\ln(\cot t){dt}=K$ \begin{align} \int_{0}^{\frac{\pi}{4}}&\ln(\cos t)dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln(\sin t)dt\\ &=\frac12\int_{0}^{\frac{\pi}{4}}\ln(\cot t){dt}+\frac12\int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt \\ &= \frac12K-\frac12\int_0^{\frac{\pi}{2}}\int_0^1 \frac{x\cot^2 t}{1+x^2 \cot^2 t}dx \ dt\\ &= \frac12K-\frac\pi4\int_0^1 \frac1{1+x}dx = \frac12K-\frac{{\pi}}{4} \ln 2 \\ \end{align}