Continuity of integral function How to show that the following function is right continuous at $0$ (that is, when $a\to0+$):
$I(a) = \int_0^{\infty}\frac{\sin x}{x}e^{-ax}dx$?
I know that Lebesgue integral $I(0) = \frac{\pi}{2}$.
Source: It's a hint in a homework problem where I'm supposed to prove that $\int_{-\infty}^{+\infty}\frac{\sin(bx)}{x}dx = \pi\operatorname{sgn}(b)$.
 A: $I(a)$ uniformly converges for $a \in [0,+\infty)$ and $f(x,a)=\dfrac{\sin x}{x}e^{-ax}$ is continuous function on $[0,+\infty)\times [0,+\infty)$, so it follows that $I(a)$ is continuous function on $[0,+\infty)$. (see picture below)
To show that $I(a)$ uniformly converges for $a \in [0,+\infty)$:
First, note that integral $\displaystyle\int_{0}^{+\infty}\frac{\sin x}{x}\, dx$ converges (in $0$ limit of function in integral is $1$, and in infinity we can use Dirichlet test). By Abel test for uniform convergence we find that $I(a)$ converges uniformly for $a\in [0,+\infty).$
Because $\displaystyle\lim_{x\to 0^+}\frac{\sin x}{x}e^{-ax}=1=f(0,a),$ we define  $$f(x,a)=\begin{cases}\frac{\sin x}{x}e^{-ax},&x>0\\ 1,&x=0 \end{cases}$$ to be continuous function on $[0,+\infty)\times [0,+\infty)$.
We used next:

You can find proof of this proposition in V. A. Zorich, Mathematical Analysis II.
A: See Appendix A of the following file: http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf. 
