We want to compute the integral \(I=\int_{-\infty}^\infty dx \frac{\sin x}{x }\) in various ways.
A complex contour integration
As it is typically done, we first upgrade real valued parameter \(x\) to a complex number \(z\) and then construct the contour in Fig. 1.
Figure 1: The complex contour in which the singularity at the origin is avoided by bending the curve around it.
On the circle of radius \(\varepsilon\), \(z=\varepsilon e^{i\theta}\) where \(\theta\in [0,\pi]\). And on the large circle \(z=Re^{i\phi}\) where \(\phi \in [0,\pi]\). We can easily evaluate the following integral (in the limit \(\varepsilon\rightarrow0\) and \(R\rightarrow\infty\)): \[\begin{eqnarray} I_c&\equiv&\oint dz\frac{e^{i z}}{z}=\int_{-R}^\varepsilon dx\frac{e^{i x}}{x}+ \int_{\pi}^0\varepsilon id\theta e^{i\theta} \frac{e^{i \varepsilon e^{i\theta}}}{\varepsilon e^{i\theta}}+\int_\varepsilon^R dx \frac{e^{i x}}{x} + \int_{0}^{\pi}R id\phi e^{i\phi} \frac{e^{i R e^{i\phi}}}{R e^{i\phi}}\nonumber\\ &=&\int_{-R}^R dx\frac{e^{i x}}{x}+\int_{\pi}^0 id\theta + \int_{0}^{\pi} id\phi e^{i R e^{i\phi}}=\int_{-R}^R dx\frac{e^{i x}}{x}-i\pi. \tag{1} \end{eqnarray}\] Note that the integral over the large circle vanishes as \(R\rightarrow\infty\) since \(e^{i R e^{i\phi}}= e^{-R \sin\phi}e^{iR \cos\phi}\). Therefore, by explicit evaluation, we see that \(I_c=\int_{-\infty}^\infty dx\frac{e^{i x}}{x}+i\pi\). But, from the theory of residues, we know that the closed loop integral of a function is \(0\) if the contour does not enclose any poles. Therefore \[\begin{eqnarray} \int_{-\infty}^\infty dx\frac{e^{i x}}{x}=i\pi, \tag{2} \end{eqnarray}\] and if we take the imaginary parts of the bothsides, we get \[\begin{eqnarray} \int_{-\infty}^\infty dx\frac{\sin x}{x}=\pi. \tag{3} \end{eqnarray}\] Side note: we evaluated the integral over the inner half circle explicitly. We could also see that it would give \(i\pi\) by observing that it is half of a circle that would have enclosed the singularity at the origin. Integral over the full circle would give \(2\pi i\), and the integral over the upper-half gives \(i\pi\).
Parametric Laplace transform
One of my favorite tricks in integration is to introduce a parameter in the integrand and manipulate it to simplify the integral. Let us insert an \(\alpha\) parameter in \(\sin\): \[\begin{eqnarray} I(\alpha)=\int_{-\infty}^\infty dx\frac{\sin(\alpha x)}{x}. \tag{4} \end{eqnarray}\] Let us apply a Laplace transform with respect to \(\alpha\) to be followed by the inverse Laplace transform
\[\begin{eqnarray} I&=&\mathscr{L} ^{-1}\left[\mathscr{L} [I(\alpha)]\right]=\mathscr{L} ^{-1}\left[\int_{-\infty}^\infty dx\mathscr{L} \left[\frac{\sin(\alpha x)}{x}\right]\right]=\mathscr{L} ^{-1}\left[\int_{-\infty}^\infty dx \frac{1}{s^2+x^2} \right]\nonumber\\ &=&\mathscr{L} ^{-1}\left[\frac{1}{s}\int_{-\infty}^\infty d(x/s) \frac{1}{1+(x/s)^2} \right]=\mathscr{L} ^{-1}\left[\frac{1}{s} \arctan(x/s)\bigg\rvert_{-\infty}^{\infty}\right]=\pi\mathscr{L} ^{-1}\left[\frac{1}{s}\right]\nonumber\\ &=&\pi. \tag{5} \end{eqnarray}\]
Direct Laplace transform
Here is a reminder on the definition of the Laplace transform: \[\begin{eqnarray} F(s)=\mathscr{L} \left[f\right]=\int_0^\infty dx e^{-s x}f(x). \tag{6} \end{eqnarray}\] From the definition, we can see that we can create a \(\frac{1}{x}\) term in the integrand if we simply integrate left side from \(s\) to \(\infty\): \[\begin{eqnarray} \int_s^\infty d\tilde s F(\tilde s)=\int_0^\infty dx \left[\int_s^\infty d\tilde s e^{-\tilde s x} \right]f(x).=\int_0^\infty dx e^{-s x}\frac{f(x)}{x}. \tag{7} \end{eqnarray}\] Therefore, if we have an expression of the form \(f(x)/x\), we can transform it as \(\int_s^\infty d\tilde s F(\tilde s)\). In our case \(f(x)=\sin x\) and \(F(s)=\frac{1}{1+s^2}\). Using the property above we get
\[\begin{eqnarray} I_s=\int_0^\infty dx e^{-s x} \frac{\sin x}{x }=\int_s^\infty d\tilde s F(\tilde s)=\int_s^\infty d\tilde s \frac{1}{1+{\tilde s}^2}=\arctan \tilde s \bigg\rvert_{s}^{\infty}=\pi/2 -\arctan s. \tag{8} \end{eqnarray}\] Note that \(I_s\) at \(s=0\) is half of the integral we are looking for: \(\int_{-\infty}^\infty dx\frac{\sin x}{x}=2 \int_{0}^\infty dx\frac{\sin x}{x}\). Doubling the result at \(s=0\) yields: \[\begin{eqnarray} I=2 I_0=\pi -2\arctan(0)=\pi \tag{9} \end{eqnarray}\]
There you have it, three ways of evaluating this lovely integral.