Evaluate $\int_{-\infty}^{\infty}\frac{\sin x}{x}\mathop{}\! \mathrm dx$ I am trying to evaluate the following:
$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x}\, dx$$
My first approach was to find the antiderivative but I can't seem to express it as I have not yet learnt about $\text{Si}(x)$. I then tried replacing the $\sin(x)$ with $(e^{ix}-e^{-ix})/(2i)$ but I just ended something even more complicated. Does making it go from $0$ to $\infty$ by multiplying by $2$ help? 
Please help me in evaluating this integral.
By the way, I am familiar with substitution and integration by parts but not complex analysis or contour integration. However, if this question requires something I don't already know, I am willing to try and understand it. 
Thanks. 
 A: 1. If you are familiar with Dirac Delta$$
\int_{-\infty}^{\infty}{\sin(x) \over x}\,{\rm d}x
=
\int_{-\infty}^{\infty}\left({1 \over 2}\,\int_{-1}^{1}{\rm e}^{{\rm i}kx}\,{\rm d}k\right)
\,{\rm d}x
=
\pi\int_{-1}^{1}{\rm d}k
\int_{-\infty}^{\infty}{{\rm d}x \over 2\pi}\,{\rm e}^{{\rm i}kx}
=
\pi\int_{-1}^{1}{\rm d}k\,\delta(k) = \pi
$$
2. Trick Calculus way
$$\begin{align*}
\int_{-\infty}^{\infty} \frac{\sin x}{x} \; dx
&= 2 \int_{0}^{\infty} \frac{\sin x}{x} \; dx \\
&= 2 \int_{0}^{\infty} \sin x \left( \int_{0}^{\infty} e^{-xt} \; dt \right) \; dx \\
&= 2 \int_{0}^{\infty} \int_{0}^{\infty}  \sin x \, e^{-tx} \; dx dt \\
&= 2 \int_{0}^{\infty} \frac{dt}{t^2 + 1} \\
&= \vphantom{\int}2 \cdot \frac{\pi}{2} = \pi.
\end{align*}$$
You would love also Feynman differentiation under the integral sign
A: Here's an approach I enjoy; maybe it's outwith the scope of your endeavour, so this may not pose as an appropriate answer
notice
$$\int \limits_0^\infty e^{-xy} \sin x dy  = \frac{\sin x}{x}$$
Hence,
\begin{align}
\int \limits_{-\infty}^{+\infty}\frac{\sin x}{x}dx
&=\int \limits_{-\infty}^{+\infty}\left(\int \limits_0^\infty e^{-xy} \sin x dy \right)dx \\
&= \int \limits_{-\infty}^{+\infty}\left(\int \limits_0^\infty e^{-xy} \sin x dx \right)dy \qquad \text{(Change order of integration)} \\
&=\int \limits_{-\infty}^{+\infty}\left(\frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2} \right)\Biggr \rvert_0^\infty dy \qquad \text{(Integrate inner brackets by parts)} \\
&=\int \limits_{-\infty}^{+\infty}\frac{1}{1+y^{2}}dy \\
&= \pi
\end{align}
