Proving an Integral Identity with Increasing Bounds How can I show that 
$$ \lim_{A\rightarrow \infty} \int_0^A \frac{\sin(x)}{x}dx\;=\;\frac{\pi}{2}?$$
I know that can use the fact that, for $x>0$, 
$$x^{-1}\;=\;\int_0^\infty e^{-xt}dt$$
but I'm not sure how to begin.
 A: Existence:
If you want your solution/elaboration to be as rigorous as it gets, it can be firstly proven that the limit indeed exists.
Start off by integration by parts, as:
$$\lim_{A \to \infty} \int_0^A \frac{\sin(x)}{x}\mathrm{d}x = \int_0^\infty \frac{\sin(x)}{x}\mathrm{d}x$$
By integration by parts, it is derived that:
$$\int_0^A \frac{\sin x}xdx=\left.\frac{1-\cos x}x\right|_0^A-\int_0^A\frac{1-\cos x}{x^2}dx$$
Using now the following simple limits 
$$\lim_{x \to 0} \frac{1-\cos(x)}{x} = 0, \; \lim_{x \to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}$$
and the fact that 
$$\int_1^\infty \frac{1}{x^2}\mathrm{d}x = 1 < + \infty$$
the existence can be straight-forwardly deriven, taking into account the comparison test with 0 \leq 1 - \cos(x) \leq 2.
Showing that it is = $\pi/2$:
To prove that it is equal to $\frac{\pi}{2}$, we will work on the following analytic expression, as:
$$1 +2\sum_{k=1}^n\cos kx=\frac{\sin \left(n+\frac 1 2\right)x}{\sin\frac x  2}$$ $$\Rightarrow$$
$$\int_0^\pi \left(1 +2\sum_{k=1}^n\cos kx\right)\mathrm{d}x = \int_0^\pi\frac{\sin \left(n+\frac 1 2\right)x}{\sin\frac x  2}\mathrm{d}x$$
$$\Rightarrow$$
$$\int_0^\pi\frac{\sin \left(n+\frac 1 2\right)x}{\sin\frac x  2}dx = \pi$$
Note that we yield that result, because: $$\int_0^\pi \cos k x \mathrm{d}x = 0 \implies \sum_{k=1}^n \int_0^\pi \cos kx \mathrm{d}x = 0$$
Moving on, the function $$f(x) = \frac{2}{x}-\frac{1}{\sin\frac x 2}$$ is continuous on $[0,\pi]$. Thus, by the Riemann-Lebesgue Lemma, $$\mathop {\lim }\limits_{n \to \infty } \int_0^\pi  {\sin \left( {n + \frac{1}{2}} \right)x\left( {\frac{2}{x} - {{\left( {\sin \frac{x}{2}} \right)}^{ - 1}}} \right)dx}  = 0$$
From here on, it can be shown that: $$\mathop {\lim }\limits_{n \to \infty } \int_0^\pi  {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{x}dx}  = \frac{\pi }{2}$$
But that ultimately yields: $$\mathop {\lim }\limits_{n \to \infty } \int_0^{\pi \left( {n + \frac{1}{2}} \right)} {\frac{{\sin x}}{x}dx}  = \frac{\pi }{2} \xrightarrow{n \to \infty} \int_0^\infty \frac{\sin(x)}{x}\mathrm{d}x = \frac{\pi}{2}$$
