Definite Integral from 0 to Quarter Pi of Logarithm of One plus Tan x

Theorem

$\ds \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x = \frac \pi 8 \ln 2$

$\ds \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x$

$=$

$\ds \int_0^{\pi/4} \map \ln {1 + \map \tan {\frac \pi 4 - x} } \rd x$

$\ds $

$=$

$\ds \int_0^{\pi/4} \map \ln {1 + \frac {\tan \frac \pi 4 - \tan x} {1 + \tan \frac \pi 4 \tan x} } \rd x$

$\ds $

$=$

$\ds \int_0^{\pi/4} \map \ln {1 + \frac {1 - \tan x} {1 + \tan x} } \rd x$

$\ds $

$=$

$\ds \int_0^{\pi/4} \map \ln {\frac 2 {1 + \tan x} } \rd x$

$\ds $

$=$

$\ds \int_0^{\pi/4} \ln 2 \rd x - \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x$

$\ds $

$=$

$\ds \frac \pi 4 \ln 2 - \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x$

Therefore:

$\ds 2 \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x = \frac \pi 4 \ln 2$

giving:

$\ds \int_0^{\pi/4} \map \ln {1 + \tan x} \rd x = \frac \pi 8 \ln 2$

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.109$