Complex Integral? Saddle Point Method? $$I(k) = \int_{0}^{∞} \frac{\sin(kx)}{x} dx$$
I have no idea how to evaluate this....  but I'm guessing the method of steepest descents (saddle point method) may work.... but not sure.
 A: You can use the fact that the sinc function is the Fourier transform of a rectangle and then apply the Plancherel theorem.
A: Very roughly since the actual solution is long to type out:
You can calculate this definite integral by using a method called "contour integration," which informally I think you can describe as line integration for complex variables.
The method behind contour integration is to take a complex-valued function, integrate it along some closed curve, and then extract the real-variable integral you want from that integration. For example, you can say:
$$ I(k) = \Im \displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ikx}}{x} \ dx $$
Now you can evaluate instead $\displaystyle \int_{\gamma} \dfrac{e^{ikz}}{z} \ dz$, where $\gamma$ is the contour that runs from $-R$ to $R$ along the real axis, with a small concave down infinitesimal semi-circle at the origin, and then along the semi-circle from $R$ to $-R$.
Notice that in the limit, along the real-axis you get $\displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ikz}}{z} \ dz $. By use of Jordan's lemma (see Wikipedia), you can show that the integral along the semi-circle goes to zero.
A: Defining 
$$\forall\,0<M\in\Bbb R\;\;\;,\;\,\Gamma_M:=\{z\in\Bbb C\;:\;|z|=M\,,\,\Im(z)>0\Longrightarrow z=Me^{it}\,,\,0\leq t\leq\pi\}$$ 
we define the complex path
$$\Gamma:=[-R,-\epsilon]\cup\Gamma_\epsilon\cup[\epsilon,R]\cup\Gamma_R\,\,\,,\,\,\epsilon,R\in\Bbb R^+\,\,,\,\epsilon<<R$$
Defining now $\,\displaystyle{f(z):=\frac{e^{ikz}}{z}}\,$ and using the corollary to the lemma in this answer , we get (all the integrals are evaluated following the positive orientation):
$$Res_{z=0}(f)=\lim_{z\to 0}e^{ikz}=1\Longrightarrow\int_{\Gamma_\epsilon}f(z)\,dz=\pi i$$
And since Jordan's Lemma gives us at once $\,\displaystyle{\int_{\Gamma_R}f(z)\,dz\xrightarrow [R\to\infty]{}0}\,$ -- or directly:
$$\left|\int_{\Gamma_R}\frac{e^{ikz}}{z}dz\right|\leq\max_{|z|=R}\frac{e^{-Rk}}{R}\pi R\xrightarrow[R\to\infty]{}0\;\;\text{(as long as}\,k>0)--$$ 
we get (since $\,f(z)\,$ is holomorphic within $\,\Gamma\,$):
$$0=\oint_\Gamma f(z)\,dz\xrightarrow[R\to\infty]{}\int_{-\infty}^\infty\frac{e^{ikx}}{x}dx-\pi i$$
and taking the imaginary part above we get
$$\int_{-\infty}^\infty\frac{i\sin kx}{x}dx=\pi i\Longrightarrow \int_0^\infty\frac{\sin kx}{x}dx=\frac{\pi}{2}$$
since the integrand is an even function.
