Integrating by parts of exponential functions I read the book Classical Fields Theory (in russian), and I met an integral taken by Landau's method:

Integrating by parts, in (8.4) we have
\begin{equation*}
\frac1{2\pi} \int\limits_{-\infty}^{\infty}dk \frac{e^{ik(x - x')} - 1}{k^2} = -\frac{|x - x'|}{2}
\end{equation*}
How the modulus ($|x- x'|$) came from integral is not entirely clear to me.
 A: The convenience in notation, let denote $k = t$ and $x - x'  = x$. Now, we're left to prove that:
$$J = \frac1{2\pi} \int\limits_{-\infty}^{\infty} \frac{e^{itx} - 1}{t^2}dt = -\frac{|x|}{2}$$
Consider the following integral on the contour $C$ of 2 semicircle in the upper half plane with center at $0$ and have $R, \epsilon$ as radius:
$$I = \frac1{2\pi}\int_{C} \frac{e^{izx} - 1}{z^2}dz$$
Since it is obvious that $\vert e^{izx} - 1 \vert \leq \vert e^{izx} \vert +1 = 2$, one can see that the integral of the semicirl with radius $R$ will go to $0$. Moreover, there is no pole inside our contour:
$$I = 0 \Rightarrow J - \lim_{\epsilon \to 0}\frac1{2\pi} \int_{C_\epsilon}\frac{e^{izx} - 1}{z^2}dz $$
$$\Rightarrow J = \lim_{\epsilon \to 0}\frac1{2\pi} \int_{C_\epsilon}\frac{e^{izx} - 1}{z^2}dz = \frac{i}{2}\text{Res} \left(\frac{e^{izx} - 1}{z^2}\right) = - \frac{x}{2}$$
Note that the above is valid for $ x > 0 $. Otherwise, we will consider two semicircle in the lower half plane which will eventually yield:
$ J = \frac{x}{2}$.
Hence:
$$J = - \frac{\vert x \vert }{2}$$
