Clean Limit Proof While attempting to solve $\int_0^\infty \frac{\sin x}{x} dx$ using Differentation Under the Integral Sign, I have stumbled across the follow limit:
$$\lim_{a \to \infty}\int _0^{\infty}\frac{x\sin \left(ax\right)}{x^2+1}dx \tag{1}$$
Now, this should go to $0$, but I have been struggling to show this cleanly without resorting to Complex Analysis. I have managed to squeeze out a proof using Integration by Parts and letting $u=\frac{x}{x^2+1}$ to get
$$\lim_{a \to \infty}\int _0^{\infty}\frac{x\sin \left(ax\right)}{x^2+1}dx =\lim_{a \to \infty}\frac{1}{2a}\int_{-\infty}^{\infty}\frac{\left(x^2-1\right)\cos \left(ax\right)}{\left(x^2+1\right)^2}dx$$
All that is left is to note that $-1 < \cos(ax) < 1$ and to apply the squeeze theorem. However, I am seeking alternative proofs that are clean and straightforward. Another way I could potentially go about this is by noting
$$\int_0^{\infty}\frac{x\sin \left(ax\right)}{x^2+1}dx = \sum_{n=0}^\infty\left(\int_{2n\pi/a}^{(2n+1)\pi/a}\frac{x\sin \left(ax\right)}{x^2+1}dx\;- \int_{(2n+1)\pi/a}^{(2n+2)\pi/a}\frac{x\sin \left(ax\right)}{x^2+1}dx\right)$$
I could now bound the difference between the two integrals; however, this seems even more tedious than my first attempt.  
What other real analysis methods can be used to evaluate (1) cleanly and efficiently?
 A: This is an extension of the Riemann-Lebesgue lemma to a uniformly convergent improper integral. 
Note that $f(x) = x/(x^2 + 1)$ is bounded and uniformly continuous on $[0,\infty)$ since $f(x) \to 0$ as $x \to \infty$.
The improper integral converges uniformly for $a$ in  any interval $[\eta,\infty)$ with $\eta >0$ since
$$\left|\int_0^c \sin(ax) \, dx \right| = \frac{|1 - \cos(ac)|}{a} < \frac{2}{\eta},$$
is uniformly bounded and $f(x) \to 0$ as $x \to \infty$ uniformly (with respect to $a$) and eventually monotonically .
Given $\epsilon > 0$ there exists $R(\epsilon) > 0$ independent of $a$ such that
$$\left|\int_{R(\epsilon)}^\infty f(x) \sin(ax) \, dx \right| < \frac{\epsilon}{3}.$$
Take a partition $(x_0, x_1, \ldots , x_n)$ of $[0,R(\epsilon)]$. Choose the number of points where the partition is so fine that $|f(x) - f(y)| < \epsilon/(3 R(\epsilon))$ for $x,y \in [x_{j-1},x_j].$  
We have
$$\left|\int_0^\infty f(x) \sin(ax) \, dx  \right| \leqslant \left|\int_0^{R(\epsilon)} f(x) \sin(ax) \, dx  \right| + \left|\int_{R(\epsilon)}^\infty f(x) \sin(ax) \, dx  \right| \\ \leqslant \left|\int_0^{R(\epsilon)} f(x) \sin(ax) \, dx  \right| + \frac{\epsilon}{3} \\ = \left|\sum_{j=1}^n \int_{x_{j-1}}^{x_j}(f(x) - f(x_j)) \sin(ax) \, dx +  \sum_{j=1}^n \int_{x_{j-1}}^{x_j}f(x_j) \sin(ax) \, dx\right| + \frac{\epsilon}{3}$$
It follows by the triangle inequality that
$$\left|\int_0^\infty f(x) \sin(ax) \, dx  \right| \\ \leqslant \sum_{j=1}^n \int_{x_{j-1}}^{x_j} |f(x) - f(x_j)| |\sin(ax)| \, dx +  \sum_{j=1}^n \left|f(x_j)\int_{x_{j-1}}^{x_j}\sin(ax) \, dx\right| + \frac{\epsilon}{3} \\ \leqslant \frac{\epsilon}{3 R(\epsilon)} R(\epsilon) + \frac{2 M n}{a} + \frac{\epsilon}{3}$$
where $M = \sup_{0 \leqslant x < \infty} f(x)$.  
Since the number of partition points $n$ depends on $\epsilon$ and is independent of $a$ we can find $A(\epsilon) > 0$ such that if $a > A(\epsilon)$ then $|I_a| < \epsilon$. This proves that $\lim_{a \to \infty} I_a = 0$.
A: By integration by parts:
$$ \int_{0}^{+\infty}\frac{x\sin(ax)}{1+x^2}\,dx =\frac{1}{a}\int_{0}^{+\infty}\frac{x^2-1}{(1+x^2)^2}(1-\cos(ax))\,dx\tag{1}$$
where the RHS of $(1)$ is an absolutely convergent integral, $\frac{x^2-1}{(1+x^2)^2}\in L^1(\mathbb{R}^+)$.
The Laplace transform of $1-\cos(ax)$ is given by $\frac{a^2}{s(a^2+s^2)}$ and the inverse Laplace transform of $\frac{x^2-1}{(1+x^2)^2}$ is given by $s\cos(s)$, hence:
$$ \int_{0}^{+\infty}\frac{x\sin(ax)}{1+x^2}\,dx = \int_{0}^{+\infty}\frac{a\cos(s)}{(a^2+s^2)}\,ds = \int_{0}^{+\infty}\frac{\cos(au)}{1+u^2}\,du\tag{2} $$
and by the dominated convergence theorem:
$$ \lim_{a\to 0}\int_{0}^{+\infty}\frac{x\sin(ax)}{1+x^2}\,dx = \lim_{a\to 0}\int_{0}^{+\infty}\frac{\cos(ax)}{1+x^2}\,dx = \int_{0}^{+\infty}\frac{dx}{1+x^2}=\color{red}{\frac{\pi}{2}}.\tag{3}$$
By the residue theorem $\int_{0}^{+\infty}\frac{\cos(ax)}{1+x^2}\,dx=\frac{\pi}{2}e^{-|a|}$, hence
$$ \lim_{a\to \infty}\int_{0}^{+\infty}\frac{x\sin(ax)}{1+x^2}\,dx = \color{red}{0}.\tag{4}$$
