Evaluating $I(s)=\int_0^{\frac{\pi}{2}}\ln(1+s\tan(\theta))\cot(\theta)d\theta$ Recently I've encountered this interesting integral$$I(s)=\int_0^{\frac{\pi}{2}}\ln(1+s\tan(\theta))\cot(\theta)d\theta=\int_0^{\infty}\frac{\ln(1+sx)}{x}\frac{dx}{x^2+1}$$
I was wondering if this integral has a closed form in terms of polylogarithms or other special functions and I would be grateful if someone could provide the answer or a method for finding it.
 A: To evaluate your integral I will do this by using Feynman's trick of differentiating under the integral sign.
Let
$$I(s) = \int_0^{\frac{\pi}{2}} \ln (1 + s \tan \theta ) \cot \theta \, d\theta, \qquad s > 0.$$
Note that $I(0) = 0$. Differentiating with respect to $s$ we have
$$I'(s) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{1 + s \tan \theta}.$$
To find the resulting integral, a $t$-substitution of $t = \tan \frac{\theta}{2}$ can be used. Here $d\theta = 2/(1 + t^2) \, dt$ and $\tan \theta = 2t/(1 - t^2)$. Thus
\begin{align}
I'(s) &= 2 \int_0^1 \frac{1 - t^2}{(1 + t^2)(1 + 2st - t^2)} \, dt\\
&= 2 \int_0^1 \left [\frac{1}{1 + s^2} \cdot \frac{1 - st}{1 + t^2} + \frac{s}{1 + s^2} \cdot \frac{t - s}{t^2 - 2st - 1} \right ] \, dt\\
&= \frac{2}{1 + s^2} \int_0^1 \frac{dt}{1 + t^2} - \frac{s}{1 + s^2} \int_0^1 \frac{2t}{1 + t^2} \, dt + \frac{s}{1 + s^2} \int_0^1 \frac{2t - 2s}{t^2 - 2st - 1} \, dt\\
&= \frac{2}{1 + s^2} \big{[}\tan^{-1} t \big{]}_0^1 -\frac{s}{1 + s^2} \big{[} \ln (1 + t^2) \big{]}_0^1 + \frac{s}{1 + s^2} \big{[} \ln |t^2 - 2st - 1| \big{]}_0^1\\
&= \frac{\pi}{2(1 + s^2)} + \frac{s \ln s}{1 + s^2}.
\end{align} 
Integrating up with respect to $s$:
\begin{align}
I(s) &= \frac{\pi}{2} \int \frac{ds}{1 + s^2} + \frac{1}{2} \int \frac{2s}{1 + s^2} \cdot \ln s \, ds\\
&= \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) - \frac{1}{2} \int \frac{\ln (1 + s^2)}{s} \, ds, \tag1
\end{align}
where in the second of the integrals integration by parts has been used. The final integral appearing in (1) can be found in terms of the dilogarithm $\operatorname{Li}_2 (z)$. To find this set: $-x = s^2$, then $ds = -dx/(2\sqrt{-x})$ and yields
$$\int \frac{\ln (1 + s^2)}{s} ds = \frac{1}{2} \int \frac{\ln (1 - x)}{x} \, dx = - \frac{1}{2} \operatorname{Li}_2 (x) + C = -\frac{1}{2} \operatorname{Li}_2 (-s^2) + C,$$
where the integral definition for the dilogarithm has been used. Thus (1) becomes
$$I(s) = \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) + \frac{1}{4} \operatorname{Li}_2 (-s^2) + C.$$
To find the constant $C$, as $I(0) = 0$, we see that $C = 0$. Thus
$$I(s) = \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) + \frac{1}{4} \operatorname{Li}_2 (-s^2), \qquad s > 0.$$
A: Assuming $s \in \mathbb{R} \wedge s > 0$, Mathematica gives:
$$I(s) = \frac{1}{24} \left(-6 \text{Li}_2\left(-\frac{1}{s^2}\right)+12 \log (s)
   \log \left(s+\frac{1}{s}\right)-12 \pi  \cot ^{-1}(s)+5 \pi ^2\right)$$
