Evaluating $\lim_{b\to\infty} \int_0^b \frac{\sin x}{x}\, dx= \frac{\pi}{2}$ 
Possible Duplicate:
Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? 

Using the identity $$\lim_{a\to\infty} \int_0^a e^{-xt}\, dt = \frac{1}{x}, x\gt 0,$$ can I get a hint to show that $$\lim_{b\to\infty} \int_0^b \frac{\sin x}{x} \,dx= \frac{\pi}{2}.$$
 A: Hint:
$$\begin{align} \lim_{b\to \infty}\int_{0}^{b}\frac{\sin x}{x}dx &= \lim_{a,b\to \infty}\int_{0}^{b}\int_{0}^{a}e^{-xt}dt\sin x dx\\& = \lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}e^{-xt}\frac{e^{ix}-e^{-ix}}{2i} dx \\&=\lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}\frac{e^{-(t-i)x}-e^{-(i+t)x}}{2i} dx\end{align}$$.
A: Too long for a comment


*

*Laplace transform:
$$\mathcal{L}\left[ \frac{f(x)}{x} \right] = \int_{0}^{\infty} \frac{f(x)}{x} e^{-yx}\, dx = \int_{y}^{\infty} \mathcal{L}\left[f(x)\right]\, ds $$

*Identity:
$$ y = 0 \implies e^{-yx} = 1$$

*Laplace transform:
$$ \mathcal{L}\left[\sin(x)\right] = \frac{1}{1+s^2}$$

*Integration:
$$ \int \frac{1}{1+s^2}\, ds = \tan^{-1}(s)$$

*Trig
$$ \tan^{-1}(\infty) = \frac{\pi}{2}$$
A: The usual procedure is as follows:
$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_0^\infty e^{-st}\frac {\sin t} {t}dt $$
We have that for any $f(t)$ such that the transform exists
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_0^\infty f(t)\frac {e^{-st}} {t}dt $$
But
$$\frac {e^{-st}} {t}=\int\limits_s^\infty e^{-mt}dm$$
Under appropriate conditions we can exchange the order of the integrands and put
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty \int\limits_0^\infty f(t) e^{-mt}dm dt $$
This means
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty F(t) dt $$
where $F$ is the transform of $f$. Using this with $\sin t$ gives
$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$
$$\int\limits_0^\infty e^{-st} \frac{\sin t}{t} dt=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$
Taking $s \to 0$
$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\frac{\pi}{2} $$
For the last step, you need to prove that
$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\int\limits_0^\infty  \frac{\sin t}{t}dt $$
I know you can use the dominated convergence theorem (which is not in my personal stash), and maybe some other theorems, but I'm unable to prove it, though I know it is legitimate (the exponential usually makes things work.) 
