Logarithmic Dilemma

Relevant wiki: Catalan's constant

$\begin{aligned} I & = \int_1^\frac \pi 4 \ln (\tan x) \ dx & \small \color{#3D99F6} \text{Let }u = \tan x \implies du = \sec^2 x \ dx \\ & = \int_{\tan 1}^1 \frac {\ln u}{u^2+1} \ du \\ & = \int_{\tan 1}^1 \frac {\ln u}{(u+i)(u-i)} du \\ & = \frac i2 \left(\int_{\tan 1}^1 \frac {\ln u}{u+i} du - \int_{\tan 1}^1 \frac {\ln u}{u-i} du \right) & \small \color{#3D99F6} \text{Let }v = u \pm i \implies du - dv \\ & = \frac i2 \left(\int_{\tan 1+i}^{1+i} \frac {\ln (v-i)}v dv - \int_{\tan 1-i}^{1-i} \frac {\ln (v+i)}v dv\right) \\ & = \frac i2 \left(\int_{\tan 1+i}^{1+i} \frac {\ln (iv+1)}v dv - \int_{\tan 1+i}^{1+i} \frac {\ln i}v dv - \int_{\tan 1-i}^{1-i} \frac {\ln (1-iv)}v dv - \int_{\tan 1-i}^{1-i} \frac {\ln i}v dv \right) & \small \color{#3D99F6} \text{Let }w = \mp iv \implies dw = \mp i \ dv \\ & = \frac i2 \left(\int_{1-i\tan 1}^{1-i} \frac {\ln (1-w)}w dw - \ln i \ln v \bigg|_{\tan 1+i}^{1+i} - \int_{1+i\tan 1}^{1+i} \frac {\ln (1-w)}w dw - \ln i \ln v \bigg|_{\tan 1-i}^{1-i} \right) \\ & = \frac i2 \left(- \text{Li}_2 (w) \bigg|_{1-i\tan 1}^{1-i} - \frac {\pi i} 2 \ln \frac {1+i}{\tan 1+i} + \text{Li}_2 (w) \bigg|_{1+i\tan 1}^{1+i} - \frac {\pi i} 2 \ln \frac {1-i}{\tan 1-i} \right) & \small \color{#3D99F6} \text{where } \text{Li}_2 (z) \text{ is the dilogarithm function.} \\ & = \frac i2 \left({\color{#3D99F6}- \text{Li}_2 (1-i)} + {\color{#D61F06}\text{Li}_2 (1-i\tan 1)} + {\color{#3D99F6}\text{Li}_2 (1+i)} - {\color{#D61F06}\text{Li}_2 (1+i\tan 1)} + \frac {\pi i} 2 \ln \frac {\tan^2 1 + 1}2 \right) \\ & = \frac i2 \left({\color{#3D99F6}2iG + \frac {\pi i}2\ln 2} - {\color{#D61F06}3.67241i} + \frac {\pi i} 2 \ln \frac {\tan^2 1 + 1}2 \right) & \small \color{#3D99F6} \text{where }G \approx 0.9159656 \text{ is the Catalan's constant.} \\ & = -G + \frac {3.67241}2 - \frac \pi 4 \ln (\tan^2 1 + 1) \\ & = \boxed{-0.04678} \end{aligned}$