Evaluate $\int_{0}^{\frac{\pi}{4}}\ln(\cos(t))dt$ $$\int_{0}^{\frac{\pi}{4}}\ln(\cos(t))dt=\frac{-{\pi}\ln(2)}{4}+\frac{K}{2}$$
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. It being $\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k}}{k}$
Giving $\int_{0}^{\frac{\pi}{4}}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}e^{-2kit}}{k}$
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?. $K=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx .916$
Your learned input is appreciated. 
 A: I dont know what a Catalan number is but the real part of the integral gets you the expression you are seeking. 
You have 
$$\int_{0}^{\frac{\pi}{4}}\ln(e^{it}+e^{-it})dt= i(2n\frac{\pi^2}{4}+\frac{\pi^2}{16}) + \int_{0}^{\frac{\pi}{4}}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}e^{-2kit}}{k}$$
Here $n$ just appears because of the multivalued $\ln (e^{it})$ term but you can ignore it as they are imganiary. 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 
$$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2(2n+1)^{2}}$$
$$=\frac{K}{2}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t
     =-\,{\pi\ln\pars{2} \over 4} + {K \over 2}:\ {\large ?}}$ where
$\ds{K \equiv
     \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} \approx 0.9160}$ 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
\\[3mm]&=-\,{\pi\ln\pars{2} \over 4} + \half\int_{0}^{\pi/4}\ln\pars{\cot\pars{t}}\,\dd t
+\int_{0}^{\pi/4}\ln\pars{2\cos\pars{t} \over \cot^{1/2}\pars{t}}\,\dd t
\end{align}
  Since ( see this link ) $\ds{K = \int_{0}^{\pi/4}\ln\pars{\cot\pars{t}}\,\dd t}$:
  $$
\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t}
=-\,{\pi\ln\pars{2} \over 4} + {K \over 2}
+ \half
\color{#00f}{\int_{0}^{\pi/4}\ln\pars{4\cos^{2}\pars{t} \over \cot\pars{t}}\,\dd t}
\tag{1}
$$

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}
=\int_{0}^{\pi/4}\ln\pars{4\sin\pars{t}\cos\pars{t}}\,\dd t
=\int_{0}^{\pi/4}\ln\pars{2\sin\pars{2t}}\,\dd t
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{2\sin\pars{t}}\,\dd t
={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
\\[3mm]&={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
\\[3mm]&={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}}
\\[3mm]&={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}}} = \color{#00f}{\large 0}
\quad\mbox{since}\quad\Gamma\pars{1} = 1.\tag{2}
\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:
  $$
\color{#00f}{\large\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t
=-\,{\pi\ln\pars{2} \over 4} + {K \over 2}}
$$

