What methods can be used to solve $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $

Question

I'm seeking methods to solve the following definite integral:

$$ I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $$

Answer 1

The method I took was:

First make the substitution $t = \tan(x)$

$$ I = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt $$

Now, let

$$ I\left(\omega\right) = \int_{0}^{\infty} \frac{\arctan(\omega t)}{t\left(1 + t^2\right)} \:dt $$

Thus,

\begin{align} \frac{dI}{d\omega} &= \int_{0}^{\infty} \frac{t}{t\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \:dt \\ &= \int_{0}^{\infty} \frac{1}{\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \\ &= \frac{1}{\omega^2 - 1} \int_{0}^{\infty}\left[\frac{\omega^2}{\left(1 + \omega^2t^2\right)} - \frac{1}{\left(1 + t^2\right)}\right]dt \\ &= \frac{1}{\omega^2 - 1} \left[\omega\arctan(\omega t) - \arctan(t) \right]_{0}^{\infty} \\ &= \frac{1}{\omega^2 - 1} \left[\omega\frac{\pi}{2} - \frac{\pi}{2}\right]\\ &= \frac{1}{\omega + 1}\frac{\pi}{2} \end{align}

Hence,

$$ I(\omega) = \int \frac{1}{\omega + 1}\frac{\pi}{2}\:d\omega = \frac{\pi}{2}\ln|\omega + 1| + C$$

Setting $\omega = 0$ we find:

$$I(0) = C = \int_{0}^{\infty} \frac{\arctan(0 \cdot t)}{t\left(1 + t^2\right)}\:dt = 0 $$

Thus,

$$ I(\omega) = \frac{\pi}{2}\ln|\omega + 1| $$

And finally,

$$I(1) = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt =\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \frac{\pi}{2}\ln(2)$$