Showing limit exists for integral of $(\sin x)/x$ Suppose we have a function like 
$$f(x)=\dfrac{\sin x}{x}$$
Consider
$$\lim_{t\rightarrow\infty}\int_1^tf(x)dx$$
According to Wolfram, this limit exists but has a value that cannot be expressed in a simple form. Clearly we will not be able to compute it by hand, but is it possible to show that the limit exists at all?
 A: Well, the limit is $$\int_0^\infty \frac{\sin t}tdt-\int_0^1\frac{\sin t}t dt$$ which is what Wolphram gives. Do you know how to show that $$\int_0^\infty \frac{\sin t}tdt$$ exists? This is a famous integral, so you might easily find literature about it. Here I provide a way to do so, and show it equals $\dfrac \pi 2$.
A: I have a solution using infinite series to solve this...
(Ah..it was mentioned by  @Callus "15mins" before in the above comments while I was still struggling with Latex...)
Note that since 
$$
\text{$\forall $a,b$\in $ R b$>$a$\geqslant $1, $\exists $}\int_a^b \frac{\sin (x)}{x} \, dx
$$
We can reduce the limit to this:
$$
\text{$\quad $ $\exists $ }
\lim_{t\rightarrow \infty } \int_1^t \frac{\sin (x)}{x} \, dx
\text{ $\Leftrightarrow \exists $}
\underset{k\rightarrow \infty }{\text{  }\lim }\int_{\pi }^{(2k+1)\pi } \frac{\sin (x)}{x} \, dx
$$
This allows us to split the integral into parts and change the problem into the sum of an alternating series.
$\underset{k\rightarrow \infty }{\text{  }\lim }\int_{\pi }^{(2k+1)\pi } \frac{\sin (x)}{x} \, dx$
$
\text{= }
\int_{\pi }^{2\pi } \frac{\sin (x)}{x} \, dx+\int_{2\pi }^{3\pi } \frac{\sin (x)}{x} \, dx+\text{...}+\int_{2\text{k$\pi $}}^{(2k+1)\pi } \frac{\sin (x)}{x} \, dx
\text{+...}
$
$
=-\int _{\pi }^{2\pi }\left|\frac{\sin (x)}{x}\right|dx+\int _{2\pi }^{3\pi }|\frac{\sin (x)}{x}|dx-\text{...}-\int _{(2k-1)\pi }^{2\text{k$\pi $}}|\frac{\sin (x)}{x}|dx+\int _{2\text{k$\pi $}}^{(2k+1)\pi }|\frac{\sin (x)}{x}|dx
$
$
=
\sum _{m=1}^{\infty } (-1)^m\int _{\text{m$\pi $}}^{(m+1)\pi }\left|\frac{\sin (x)}{x}\right|dx
$
$
=
\sum _{m=1}^{\infty } (-1)^m\alpha _m, \alpha _m=\int _{\text{m$\pi $}}^{(m+1)\pi }\left|\frac{\sin (x)}{x}\right|dx
$
Then it's easy to see that $\alpha _m>\alpha _{m+1}$
and $\lim_{m\rightarrow \infty } \alpha _m=0.$
Thus it passes the "Leibniz Test"(http://en.wikipedia.org/wiki/Alternating_series_test) and therefore it converges.
