Continuity of Laplace transform I'm trying to prove the Riemann integral $$\int_0 ^ \infty \frac{\sin(x)} x \ dx = \frac{\pi}{2}$$ using tools that I, as a statistician, am likely not to forget. My general approach is 


*

*Show that the Laplace transform $\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$ is $-\arctan(t) + \frac \pi 2$ for $t \in (0, \infty)$ by differentiating under the integral (using dominated convergence stuff since the Lebesgue integral exists here) and solving the resulting differential equation. 

*Argue that $\psi(t)$ is continuous at $0$. 


I've hit a snag at step 2. So, under what conditions do we have that the Laplace transform of a function is continuous at $t = 0$? Any trivial garuntees that it should be for $\frac {\sin x} x$ in particular? I feel like this would be easier if the function was Lebesgue integrable since when I'm trying to bound $|\psi(0) - \psi(t)|$ I am having a hard time controlling the tail of the integral of $\psi(t)$ for $t$ uniformly in $t$ near $0$. 
I'm also finding this somewhat interesting since it looks like $\frac{\sin x} x$ is sitting on a boundary of sorts: we are Lebesgue integrable up to a point, then Riemann integrable, and past that not integrable($t > 0, = 0, < 0$) So a bunch of Lebesgue integrable functions are converging to a function which has an improper Riemann integral but no Lebesgue integral. And yet interchanging limits produces the correct answer anyways when calculating the integral. 
EDIT: We already have an answer below that suggests how to do this for $\frac {\sin x} x$ but I'd like to get something on the more general question of under what conditions we have right continuity of the Laplace transform at $0$. Is it always continuous provided that the function has a finite improper Riemann integral?
 A: You can get right continuity at $s=0$ along the real axis, at least: We have
$$I(s,n):=\int_{n\pi}^\infty \frac{e^{-st}\sin t}{t}\,dt=\sum_{k=n}^\infty\int_{k\pi}^{(k+1)\pi} \frac{e^{-st}\sin t}{t}\,dt$$
and use the alternating nature of the series to show that the entire sum lies between $0$ and its first term. Thus $I(s,n)\to0$ uniformly in $s\ge0$ as $n\to\infty$. On the other hand, you have unifom convergence of the integrand for $t\in[0,n\pi]$, so you don't even need to invoke Lebesgue theory to get the needed convergence.
A: You can use an alternate result which is quite nice. 
$${\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_0^\infty  {{e^{ - st}}\frac{{\sin t}}{t}dt} $$
But
$$\frac{{{e^{ - st}}}}{t} = \int\limits_s^\infty  {{e^{ - tm}}dm}   $$
So
$$\eqalign{
  & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_0^\infty  {\left( {\int\limits_s^\infty  {{e^{ - tm}}} dm} \right)\sin tdt}   \cr 
  & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_s^\infty  {\left( {\int\limits_0^\infty  {{e^{ - tm}}} \sin tdt} \right)dm}   \cr 
  & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_s^\infty  {\frac{{dm}}{{1 + {m^2}}}}  \cr} $$
This means that
$$\int\limits_0^\infty  {{e^{ - st}}\frac{{\sin t}}{t}dt}  = \int\limits_s^\infty  {\frac{{dm}}{{1 + {m^2}}}} $$
Now let $s \to 0$.
A: I would like to show my way to show that $ \int_0 ^ \infty \frac{\sin(x)} x \ dx=\frac{\pi}{2}$
$\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$
$\psi '(t) := -\int_0 ^ \infty e^{-xt} \sin(x)  \ dx$
From Laplace tables we can find sine's laplace transform=$\frac{1}{t^2+1}$  . if you don't want to use Laplace table, replace $sin(x)=\frac{e^{ix}-{e^{-ix}}}{2i}$ and find the same result easly from $ \int_0 ^ \infty e^{-ax} \ dx=\frac{1}{a}$.
$\psi '(t) := -\frac{1}{t^2+1} $
$\int \psi '(t)dt=-\int \frac{1}{t^2+1}  dt$
$\psi(t) := -arctan(t)+c $
if $\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$
then
$\psi(\infty)=\int_0 ^ \infty \frac{e^{-\infty t} \sin(x)} x \ dx =0$
if $arctan(\infty)=\frac{\pi}{2}$
 then 
$c=\frac{\pi}{2}$
$\psi(0) := \int_0^\infty \frac {sin(x)}{x}  dx=-arctan(0)+c=\frac{\pi}{2}$ 
