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Logarithmic integral with tangent

Let a \in \mathbb{R}. Evaluate the integral

    \[\mathcal{J} = \int_{0}^{a} \ln \left ( 1 + \tan a \tan x \right ) \, \mathrm{d}x\]

Solution

We distinguish cases: If a \in \mathbb{R} \mid \tan a =0 then the integral equals 0. Otherwise,

    \begin{align*} \mathcal{J} &= \int_{0}^{a} \ln \left ( 1 + \tan a \tan x \right ) \, \mathrm{d}x \\ &\!\!\!\!\!\!\overset{x \mapsto a- x}{=\! =\! =\! =\! =\!} \int_{0}^{a} \ln \left ( 1 + \tan a \tan \left ( a - x \right ) \right ) \, \mathrm{d}x \\ &= \int_{0}^{a} \ln \left ( 1 + \tan a \cdot \frac{\tan a - \tan x}{1 + \tan a \tan x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{a} \ln \left ( 1 + \frac{\tan^2 a - \tan a \tan x}{1+ \tan a \tan x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{a} \ln \left ( \frac{1 + \tan a \tan x + \tan^2 a - \tan a \tan x}{1+ \tan a \tan x} \right ) \, \mathrm{d}x \\ &= \int_{0}^{a} \ln \frac{1 + \tan^2 a}{1+ \tan a \tan x} \, \mathrm{d}x \\ &= \int_{0}^{a} \ln \left( 1 +  \tan^2 a \right) \, \mathrm{d}x - \mathcal{J} \\ &= a \ln \left( 1 +  \tan^2 a \right)  - \mathcal{J} \end{align*}

Hence

    \[\mathcal{J} = \frac{a \ln \left( 1 +  \tan^2 a \right)}{2}\]

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