Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$

Under $\tan x\to x$, one has \begin{eqnarray} &&\int_0^{\pi/4}\ln(\tan x)dx\\ &=&\int_0^1\frac{\ln x}{1+x^2}dx\\ &=&\int_0^1\ln x\sum_{n=0}^\infty(-1)^nx^{2n}dx\\ &=&-\sum_{n=0}^\infty\int_0^1(-1)^nx^{2n}\ln xdx\\ &=&-\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)^2}\\ &=&-C \end{eqnarray} where $C$ is the Catalan constant.

Another way is to use the Fourier expansion of $\log\sin x$ and $\log\cos x$ and integrate the expression term-wise. Since

$$\begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos 2kx}k-\log 2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\end{align*}$$

Therefore$$\begin{align*}I & =\int\limits_0^{\pi/4}dx\,\log\tan x\\ & =\int\limits_0^{\pi/4}dx\,\log\sin x-\int\limits_0^{\pi/4}dx\,\log\cos x\\ & =-\sum\limits_{k\geq1}\int\limits_0^{\pi/4}dx\,\frac {\cos 2kx}k+\sum\limits_{k\geq1}\int\limits_0^{\pi/4}dx\, (-1)^k\frac {\cos 2kx}k\end{align*}$$

Now integrate within the limits to get$$\begin{align*}I & =\frac 12\sum\limits_{k\geq1}\frac {(-1)^k\sin\left(\frac {\pi k}2\right)}{k^2}-\frac 12\sum\limits_{k\geq1}\frac {\sin\left(\frac {\pi k}2\right)}{k^2}\\ & =-\frac 12\sum\limits_{k\geq1}\frac {(-1)^k}{(2k-1)^2}-\frac 12\sum\limits_{k\geq1}\frac {(-1)^k}{(2k-1)^2}\\ & =-G\end{align*}$$Where $G$ is Catalan's Constant.