real analysis - Computing $lim_{epsilon rightarrow 0} int_0^infty frac{sin x}{x} arctan{frac{x}{epsilon}}dx$

I'm not exactly sure how to get started computing the limit of the improper Riemann integral

$$\lim_{\epsilon \rightarrow 0} \int_0^\infty \frac{\sin x}{x} \arctan\left(\frac{x}{\epsilon}\right)dx.$$

Using the result that $\int_0^\infty \frac{\sin x}{x} dx = \pi/2$, is there a way to interchange the limit and the integral to get $\pi^2/4$?

Answer

By the dominated convergence theorem

$$\lim_{\epsilon \to 0} \int_0^\pi \frac{\sin x}{x} \arctan\frac{x}{\epsilon}\,dx=\frac{\pi}{2}\int_0^\pi \frac{\sin x}{x}\,dx.$$
Now
$$\int_\pi^\infty \frac{\sin x}{x}\,\arctan\frac{x}{\epsilon}\,dx=\frac{\pi}{2}\int_\pi^\infty \frac{\sin x}{x}\,dx+\int_\pi^\infty \frac{\sin x}{x}\Bigl(\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr)\,dx.$$
Let's stimate the second integral:
$$\Bigl|\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr|=\int_{x/\epsilon}^\infty\frac{dt}{1+t^2}\le\frac{\epsilon}{x},$$
and
$$\int_\pi^\infty \Bigl|\frac{\sin x}{x}\Bigl(\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr)\Bigr|\,dx\le\epsilon\int_\pi^\infty\frac{|\sin x|}{x^2}\,dx.$$