Cannot understand use Feymann's trick to calculate improper integral We can use Feymann trick to calculate $\int_0^\infty \frac{\sin x}{x}dx$. First, let
$$ I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}dx, a> 0.$$
It is easy to see
$$I'(a)=-\int_0^\infty e^{-ax}\sin xdx=\frac{e^{-ax}}{a^2+1}(a\sin x+\cos x)\big|_0^\infty=-\frac{1}{a^2+1}.$$
Thus
Integrating it we find that 
$$I(a)=-\arctan(a)+C$$
Letting $a\to \infty$ in reveals $\lim_{a\to \infty}I(a)=0$ and hence $C=\pi/2$.
so $I(a)=-\arctan(a)+\frac{\pi}{2}$
Finally, 
$\lim_{a\to 0^+}I(a)=\lim_{a\to 0^+}-\arctan(a)+\frac{\pi}{2}=\frac{\pi}{2}$
so $I(0)=\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2} (*)$
I cannot understand the last step(*), why $\lim_{a\to 0^+}I(a)=I(0)$, I think we need to prove it, but how to prove?
 A: The improper integral
$$I(a) = \int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx$$
is uniformly convergent for $a \in [0,\infty)$ and, hence, $a \mapsto I(a)$ is continuous on $[0,\infty)$.  This implies $\lim_{a \to 0+} I(a) = I(0)$.
The uniform convergence follows from the Dirichlet test as $ \int_0^c \sin x \, dx$ is uniformly bounded for all $c > 0$, $a \in [0,\infty)$ and $\frac{e^{-ax}}{x} \to 0$ as $x  \to \infty$ monotonically and uniformly.
To understand why uniform convergence here implies continuity, note that
$$|I(a_1) - I(a_2)| \leqslant \int_0^c\left|(e^{-a_2x}- e^{-a_1x})\frac{\sin x}{x} \right|\, dx+ \left|\int_c^\infty e^{-a_1x}\frac{\sin x}{x}\, dx \right| \\+ \left|\int_c^\infty e^{-a_2x}\frac{\sin x}{x}\, dx \right|$$
The second and third terms on the RHS can be made smaller than $\epsilon/3$ by choosing $c$ sufficiently large (independent of $a_1,a_2$) by uniform convergence of the integral.  The first term on the RHS can be made smaller than $\epsilon/3$ by choosing choosing $a_1,a_2$ sufficiently close since the integrand is a continuous function and the integral is taken over a finite interval. 
