Evaluating $\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$ Seems to be a hard nut:  
$$I=\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$$ Any hint?
 A: It is not hard to show that
\begin{equation}\int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty, \quad \cdots \quad (1)\end{equation}
and it is also easy to show that
$$\arctan x \geq Cx \quad \text{for} \quad 0 \leq x \leq 1 \quad \cdots \quad (2)$$
for some positive constant $C > 0$. Now it is clear that these together imply
\begin{equation}\int_{0}^{\infty} \frac{\arctan \sin^2 x}{x} \; dx \geq C \int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty\end{equation}

Indeed, we first show that $(1)$ diverges. It suffices to show that
$$ \int_{2012}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty. $$
By integration by parts, we have
$$ \begin{align*}
\int_{2012}^{R} \frac{\sin^2 x}{x} \; dx
&= \left[ \frac{1}{2} - \frac{\sin 2x}{4x}\right]_{2012}^{R} + \int_{2012}^{R} \left( \frac{1}{2x} - \frac{\sin 2x}{4x^2}\right) \; dx\\
&= \frac{1}{2}\log R + O(1),
\end{align*}$$
which proves $(1)$ by letting $R\to\infty$.
Now we prove $(2)$. by examining second derivative of arc-tangent function, we find that it is concave on $[0, 1]$. Thus on this interval we have
$$\arctan x \geq (\arctan 1) x,$$
which proves $(2)$ with $C=\arctan 1 =\frac{\pi}{4}$.
