Proving interchangeability of limit and integral for $\lim_{k\to0}\int_0^{\infty}\frac{\sin x}{x(k^2x^2+1)}\mathrm{d}x$ As the title asks, I would like to ask if $\lim_{k\to0}\int_0^{\infty}\frac{\sin x}{x(k^2x^2+1)}\mathrm{d}x=\int_0^{\infty}\lim_{k\to0}\frac{\sin x}{x(k^2x^2+1)}\mathrm{d}x$. I'm sure there's some form of convergence theorem that I have to use, and highly likely that it's the DCT, but I can't think of the sequence of functions I should use. Could someone guide me on this please? Thanks!
 A: HINT:
Note that we have
$$\begin{align}
\int_0^\infty \frac{\sin(x)}{x(k^2x^2+1)}\,dx&=\int_0^\infty \left(\frac{\sin(x)}{x}-\frac{k^2x\sin(x)}{k^2x^2+1}\right)\,dx\\\\
&\overbrace{=}^{x\mapsto x/k}\frac\pi2 -\int_0^\infty \frac{x\sin(x/k)}{x^2+1}\,dx\\\\
&\overbrace{=}^{\text{IBP}}\frac\pi2 +k\underbrace{\int_0^\infty \left(\frac{1-x^2}{(1+x^2)^2}\cos(x/k)\right)\,dx}_{\text{Absolutely Convergent}}
\end{align}$$
Can you finish now?
A: The function $f: [0,\infty)\times \mathbb{R} \to \mathbb{R}$ with
$$f(x,k) = \begin{cases} \frac{\sin x}{x(k^2x+1)} , & x \neq 0 \\0, & x = 0\end{cases}$$
is continuous. For any $R> 0$, it follows that $F(k) = \int _0^Rf(x,k) \, dx$ is continuous and
$$\tag{1}\lim_{k \to 0} \int_0^R\frac{\sin x}{x(k^2x+1)}\, dx = \int_0^R \frac{\sin x }{x} \, dx$$
(Swapping the limit and the integral of a continuous function over a finite interval is permitted)
Note that
$$\left|\int_0^\infty\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^\infty\frac{\sin x}{x}\, dx   \right| = \left|\int_R^\infty\frac{\sin x}{x(k^2x+1)}\, dx + \int_0^R\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^R\frac{\sin x}{x}\, dx -  \int_R^\infty\frac{\sin x}{x}\, dx \right|\\ \leqslant \left| \int_0^R\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^R\frac{\sin x}{x}\, dx \right|+\left|\int_R^\infty\frac{\sin x}{x(k^2x+1)}\, dx\right| + \left|  \int_R^\infty\frac{\sin x}{x}\, dx\right|$$
Since the improper integral $\int_0^\infty \frac{\sin x}{x} \, dx$ is convergent, for any $\epsilon > 0$, there exists $R_1>0$ such that $ \left|  \int_R^\infty\frac{\sin x}{x}\, dx\right| < \frac{\epsilon}{3}$ for all $R > R_1$.
We also have, by the Dirichlet test for improper integrals, uniform convergence of $\int_0^\infty\frac{\sin x}{x(k^2x+1)}\, dx$ and there exists $R_2 > 0$ such that $\left|\int_R^\infty\frac{\sin x}{x(k^2x+1)}\, dx\right| < \frac{\epsilon}{3}$ for all $R > R_2$ and all $k \in \mathbb{R}$.
Hence, choosing any $R > \max(R_1,R_2)$ we have
$$\tag{2}\left|\int_0^\infty\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^\infty\frac{\sin x}{x}\, dx   \right| \leqslant  \left| \int_0^R\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^R\frac{\sin x}{x}\, dx \right|+ \frac{2\epsilon}{3}$$
By (1) it follows that there exists $\delta > 0$ (which may depend on $R$ without consequence) such that if $|k-0| < \delta$, then
$$\tag{3} \left|\int_0^R\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^R\frac{\sin x}{x}\, dx   \right| \leqslant \frac{\epsilon}{3},$$
and together with $(2)$ this implies
$$\left|\int_0^\infty\frac{\sin x}{x(k^2x+1)}\, dx - \int_0^R\frac{\sin x}{x}\, dx   \right| \leqslant \epsilon$$
Therefore,
$$\lim_{k \to 0} \int_0^\infty\frac{\sin x}{x(k^2x+1)}\, dx = \int_0^\infty \frac{\sin x }{x} \, dx$$
