Interchange of a limit and integral How would one show the following equality? Trying standard convergence theorems eg DCT (I can't find a choice of a dominating function which is also integrable) do not appear to be successful.
$$\lim_{a_n \to 0^+} \int_\mathbb{R} e^{−a_n|x|}\frac {\sin(x)}x dx=\int_\mathbb{R} \frac {\sin(x)}x dx $$
Thanks
 A: With $a > 0$ we have
$$\int_{\mathbb{R}}e^{-a|x|} \frac{\sin x}{x} \, dx = 2 \int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$\tag{*}\lim_{a \to 0+}\int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx = \int_0^\infty \lim_{a \to 0+}e^{-ax} \frac{\sin x}{x} \, dx = \int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a \geqslant 0$.  To prove this, note that for any $c > 0$
$$\tag{**}\left|\int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx  - \int_0^\infty  \frac{\sin x}{x} \, dx    \right| \\ \leqslant \left|\int_0^c e^{-ax} \frac{\sin x}{x} \, dx  - \int_0^c  \frac{\sin x}{x} \, dx    \right| + \left|\int_c^\infty e^{-ax} \frac{\sin x}{x} \, dx      \right| + \left|\int_c^\infty  \frac{\sin x}{x} \, dx      \right| $$
Since the integrand is uniformly continuous for $(x,a) \in [0,c] \times [0,b]$ for any $b > 0$, there exists $\delta > 0$ such that the first term on the RHS is less than $\epsilon/3$ when $0 < a <\delta$.  This, of course, is the justification for switching the limit and integral over a bounded interval.  By uniform convergence of the improper integral of $e^{-ax} \sin x / x$ and the convergence of the improper integral of $\sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $\epsilon/3$ as well. 
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $\xi$ with $c_1 < \xi < c_2$ such that
$$\left|\int_{c_1}^{c_2} e^{-ax} \frac{\sin x}{x} \, dx \right| = \frac{e^{-ac_1}}{c_1}\left|\int_{c_1}^{\xi}\sin x \, dx \right| = \frac{e^{-ac_1}}{c_1}\left|\cos c_1 - \cos \xi\right| \leqslant \frac{2}{c_1}$$
Now the RHS is smaller than $\epsilon >0$ by choosing $c_1 > 2/ \epsilon$ for any $a \geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
