Evaluate $\displaystyle \int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$
$\displaystyle \int_{-\infty}^{+\infty}\frac{\sin x}x\,dx = \int_{-\infty}^0 \frac{\sin x}x\,dx + \int_0^{+\infty} \frac{\sin x}x\,dx$
$\displaystyle = \lim_{a \rightarrow -\infty} \int_a^0 \frac{\sin x}x\,dx + \lim_{b \rightarrow +\infty} \int_0^b \frac{\sin x}x\,dx$
now, integrate by parts.. continue..
I don't think sin(x)/x is readily done by parts or any other elementary operation. This is a case for PH's forte, complex analysis.
You can start by using $\displaystyle \frac{e^{iz}}{z}$, because it has a simple pole at z=0 and has a residue $\displaystyle e^{i(0)}=1$
I have been studying a little CA when I have time, but don't have my sea legs yet.
$\displaystyle \frac1x=\int_0^\infty e^{-ux}\,du.$ so,
$\displaystyle \int_{ - \infty }^{ + \infty } {\frac{{\sin x}}
{x}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin x\,du} \,dx} .$
Now the remaining challenge is to compute $\displaystyle {\int_0^\infty {e^{ - ux} \sin x\,dx} }.$ (Reverse integration order.)
This is nasty applyin' integration by parts, but it's quickly considering the following:
$\displaystyle \int_0^\infty {e^{ - ux} \sin x\,dx} = \text{Im} \int_0^\infty {e^{ - (u - i)x} \,dx} = \text{Im} \frac{1}
{{u - i}} = \frac{1}
{{u^2 + 1}}.$
And finally
$\displaystyle \int_{ - \infty }^{ + \infty } {\frac{{\sin x}}
{x}\,dx} = 2\int_0^\infty {\frac{1}
{{1 + u^2 }}\,du} = \pi .$
Let $\displaystyle f(z) = \frac{e^{iz}}{z}$. Consider the square contour $\displaystyle C$ (for $\displaystyle A,B>0$ large) from $\displaystyle -A$ to $\displaystyle B$ with $\displaystyle \gamma$ be a semicircular contour from $\displaystyle -r$ to $\displaystyle r$ (for $\displaystyle r>0$ small).
Then by Cauchy's theorem*:
$\displaystyle \int_{r}^{\infty}\frac{e^{ix}}{x}dx + \int_{-\infty}^{-r} \frac{e^{ix}}{x} dx - \int_{\gamma} f(z)dz = 0$
Now make $\displaystyle r\to 0^+$ to get:
$\displaystyle \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx = \pi i$
Equate real and imaginary parts.
*)The issue of convergence is covered by Jordan's lemma.