Complex Integrals to find infinite integral on the real line. Right now I am trying to prove the following integral:
$$\int_{-\infty}^{\infty} \frac{\sin{\pi x}}{\pi x} dx = 1$$
And the idea is to use complex variables and rewrite this as part of a complex contour integral on a half circle. The contour path is described as follows. Imagine a half circle in the complex plane with positive imaginary component, with a large radius $R$ centered at zero. Call this outer circle $C_1$, and integrate counterclockwise. We then cut out of that a half-circle of small radius $\epsilon$ centered at zero. Call this inner circle $C_2$, and integrate clockwise. 
The idea is that we have a half-donut shaped figure in the complex plane (with $Im(z) > 0$) and that we can integrate the function $f(z)$ over this closed, simply connected contour path and get zero by Cauchy's Theorem. We define $f(z)$ as follows, and integrate over the path in a counterclockwise direction so that we are going in the positive $x$ direction on the real line:
$$f(z) = \frac{e^{i\pi z}}{\pi z} = \frac{\cos(\pi z)}{\pi z} + i\frac{\sin(\pi z)}{\pi z}$$
As $\epsilon \to 0$ and $R \to \infty$, the base is supposed to approach the real line, 
$$\int_{C_1} f(z) dz \to 0 \qquad\qquad
\int_{C_2} f(z) dz \to -i$$
And by Cauchy's Theorem:
$$\int_{C_1} f(z) dz +
\int_{C_2} f(z) dz + 
\int_{\epsilon < |x| < R} f(z) dz = 0$$
This would then indicate that
$$\int_{-\infty}^{\infty} f(z) dz = i$$
To show that $\int_{C_1} f(z) dz \to 0$, I was able to start with the following:
$$z = Re^{i\theta} = R\cos(\theta) + R\sin(\theta)i \qquad\qquad
dz = Rie^{i\theta} d\theta = iz d\theta$$
We can then change variables and bound it in the following way (I think this works, but not sure):
$$\bigg|{\int_{C_1} \frac{e^{i\pi z}}{\pi z} dz}\bigg| \leq
\int_{C_1} \bigg|\frac{e^{i\pi z}}{\pi z}\bigg| |dz| = 
\int_{0}^{\pi} \bigg|\frac{1}{\pi Re^{i\theta}}\bigg|\bigg|e^{-\pi R\sin(\theta)}\bigg| |R|d\theta = 
\frac{1}{\pi}\int_{0}^{\pi} e^{-\pi R\sin(\theta)} d\theta \overset{R \to \infty}{\longrightarrow} 0$$
Where I am having the most trouble here is seeing why $\int_{C_2} f(z) dz \to -i$.
 A: 
METHODOLOGY $1$:  CONTOUR INTEGRATION

From Cauchy's Integral Theorem we have
$$\begin{align}
0&=\oint\frac{e^{i\pi z}}{\pi z}\,dz\\\\
&=\int_{\epsilon\le |x|\le R}\frac{e^{i\pi x}}{\pi x}\,dx+\int_0^\pi \frac{e^{i\pi Re^{i\phi}}}{\pi Re^{i\phi}}\,i Re^{i\phi}\,d\phi-\int_0^\pi \frac{e^{i\pi \epsilon e^{i\phi}}}{\pi \epsilon e^{i\phi}}\,i \epsilon e^{i\phi}\,d\phi\\\\
&=\int_{\epsilon\le |x|\le R}\frac{e^{i\pi x}}{\pi x}\,dx+\frac i\pi\int_0^\pi e^{i\pi Re^{i\phi}}\,d\phi- \frac i\pi\int_0^\pi e^{i\pi \epsilon e^{i\phi}}\,d\phi\tag1
\end{align}$$
As $R\to \infty$, the second integral on the right hand side of $(1)$ approaches $0$ while as $\epsilon\to 0$, the third integral approaches $\pi$ (See the APPENDIX).  
Now, note that 
$$\int_{\epsilon\le |x|\le R}\frac{e^{i\pi x}}{\pi x}\,dx=\int_{\epsilon\le |x|\le R}\frac{\cos(\pi x)}{\pi x}\,dx+i\int_{\epsilon\le |x|\le R}\frac{\sin(\pi x)}{\pi x}\,dx \tag 2$$
The first integral on the right-hand side of $(2)$ is the integral of an odd function over symmetric limits.  Its value is, therefore, $0$.
Hence, we have
$$\int_{\epsilon\le |x|\le R}\frac{e^{i\pi x}}{\pi x}\,dx=i\int_{\epsilon\le |x|\le R}\frac{\sin(\pi x)}{\pi x}\,dx \tag3$$
whence letting $R\to \infty$ and $\epsilon\to0$ reveals
$$\lim_{R\to\infty}\lim_{\epsilon\to0}\int_{\epsilon\le |x|\le R}\frac{e^{i\pi x}}{\pi x}\,dx=i\int_{-\infty}^\infty \frac{\sin(\pi x)}{\pi x}\,dx\tag4$$
Putting it all together yields the coveted integral 
$$\int_{-\infty}^\infty \frac{\sin(\pi x)}{\pi x}\,dx=1$$


METHODOLOGY $2$:  FEYNMAN'S TRICK

Let $I(s)$ be the integral given by 
$$I(s)\int_0^\infty \frac{\sin(\pi x)}{\pi x}e^{-sx}\,dx\tag5$$
for $s>0$.
Since $\int_0^\infty \sin(\pi x)e^{-sx}\,dx$ converges uniformly for $s\ge \delta >0$, we can differentiate under the integral in $(5)$ to find
$$I'(s)=-\frac1\pi \int_0^\infty e^{-sx}\sin(\pi x)\,dx=-\frac{\pi }{\pi^2+s^2}\tag6$$
Integrating $(6)$ and using $\lim_{s\to \infty }I(s)=0$ reveals
$$I(s)=\frac{\pi/2-\arctan(s/\pi)}{\pi}$$
Letting $s\to 0$, we find that 
$$\int_0^\infty \frac{\sin(\pi x)}{\pi x}\,dx=\frac12$$
Exploiting the evenness of the integrand yields the expected result
$$\int_{-\infty}^\infty \frac{\sin(\pi x)}{\pi x}\,dx=1$$


APPENDIX:

Here, we evaluate the limit $\lim_{R\to\infty} \int_0^\pi e^{i\pi Re^{i\phi}}\,d\phi$.  
To proceed, note that we can write $e^{i\pi Re^{i\phi}}=e^{i\pi R\cos(\phi)}e^{-\pi R\sin(\phi)}$.  
Furthermore, note that the sine function is even about $\pi/2$ and that $\sin( \phi)\ge \frac2\pi \phi$ for $\phi\in[0,\pi/2]$.   
Hence, we have
$$\begin{align}
\left|\int_0^\pi e^{i\pi Re^{i\phi}}\,d\phi\right|\le&\int_0^\pi |e^{i\pi R\cos(\phi)}e^{-\pi R\sin(\phi)}|\,d\phi\\\\
&=\int_0^\pi e^{-\pi R\sin(\phi)}\,d\phi\\\\
&=2\int_0^{\pi/2}e^{-\pi R\sin(\phi)}\,d\phi\\\\
&\le 2\int_0^{\pi/2 }e^{-2R\phi }\,d\phi\\\\
& =2\left(\frac{1-e^{-2R\pi }}{2R}\right)\\\\
&\to 0\,\,\text{as}\,\,R\to \infty
\end{align}$$ 
A: Here is a real-analytic way to evaluate it. Call your integral $I$. A useful property of the Laplace transform is that
$$\int_0^{+\infty} f(x)g(x)\,dx=\int_0^{+\infty} \mathcal{L}\{f(x)\}(s)\,\mathcal{L}^{-1}\{g(x)\}(s)\,dx$$
Noting that $\text{sinc}\,\pi x$ is an even function, and choosing $f(x)=\sin \pi x$ and $g(x)=\frac{1}{x}$, we have that 
\begin{align}
I&=\frac{2}{\pi}\int_0^{+\infty} \frac{\pi}{\pi^2+s^2}\,ds \\ \\
&=2\left[\frac{1}{\pi}\arctan\left(\frac{s}{\pi}\right) \Biggr|_{s=0}^{+\infty}\right] \\ 
&=\frac{2}{\pi}\cdot \frac{\pi}{2} \\
&= \boxed{1}
\end{align}
