I'm reading a paper and it is claimed that for every $t\in \mathbb{R}$
$$\int_{-\infty}^{+\infty} (e^{itx}-1-\frac{itx}{1+x^2})\frac{1}{\pi x^2} dx = -|t|$$
Note that there is no problem at $0$ since we can give the integrand a value at $0$ such that this expression becomes continuous.
How can I see that this equality is true? Do I need complex analysis to see this?
We will use the fact that $\int^{\infty}_{-\infty}\frac{\sin tx}{x}dx=\pi\ \text{sgn}\ t.$ First, split the integral up into two parts and write the second part as ${\displaystyle\int}\dfrac{1}{\left(\frac{1}{x^2}+1\right)x^3}\,\mathrm{d}x$. We find that
$\displaystyle\int \dfrac{1}{x\left(x^2+1\right)}dx=-\dfrac{\ln\left(\frac{1}{x^2}+1\right)}{2}+C$. Thus, this part of the definite integral is zero.
For the first part, omitting the factor $1/\pi$ for the moment, integrate ${\displaystyle\int}\dfrac{\mathrm{e}^{\mathrm{i}tx}-1}{x^2}\,\mathrm{d}x$ by parts, to get $-\dfrac{\mathrm{e}^{\mathrm{i}tx}-1}{x}-{\displaystyle\int}-\dfrac{\mathrm{i}t\mathrm{e}^{\mathrm{i}tx}}{x}\,\mathrm{d}x.$ The first term gives zero so we are left with
${\displaystyle\int^{\infty}_{-\infty}}\dfrac{\mathrm{i}t\mathrm{e}^{\mathrm{i}tx}}{x}\,\mathrm{d}x=\displaystyle it\int^{\infty}_{-\infty}\frac{\cos tx}{x}dx-t\int^{\infty}_{-\infty}\frac{\sin tx}{x}dx.$
Using the definition of the Cauchy principal value, we may interpret the first of these integrals on the right-hand side as in $(1)$ in the above-cited link. It is then seen to be zero because the integrand is odd. So, finally, we have
$\displaystyle -t\int^{\infty}_{-\infty}\frac{\sin tx}{x}dx=-t(\pi\ \text{sgn}\ t)$. All that remains now is to multiply by $1/\pi$.
Define $g \colon \mathbb{R} \times (\mathbb{C} \setminus \{-\mathrm{i},\mathrm{i}\}) \to \mathbb{C},$ $$ g(t,z) = \begin{cases} \frac{1}{\pi z^2} \left[\mathrm{e}^{\mathrm{i} t z} - 1 - \frac{\mathrm{i}t z}{1+z^2}\right] & , \, z \neq 0 \\ - \frac{t^2}{2 \pi} &, \, z = 0\end{cases} \, .$$ Note that $g(t,0) = \lim_{z \to 0} g(t,z)$, so $g(t,\cdot)$ is continuous at the origin and decays quadratically at infinity (on the real axis). Therefore, the function $f \colon \mathbb{R} \to \mathbb{R}, \, f(t) = \int_\mathbb{R} g(t,x) \, \mathrm{d} x,$ is well-defined. We have $f(0) = 0$ as you mentioned and $g(-t,-z) = g(t,z)$ implies $f(-t) = f(t)$ for $t \in \mathbb{R}$, so $f$ is symmetric. Here are two ways to compute $f$:
Complex analysis
$g(t,\cdot)$ is meromorphic with residues $$ \operatorname{Res}(g(t,\cdot),\pm \mathrm{i}) = \lim_{z \to \pm \mathrm{i}} (z \mp \mathrm{i}) g(t,z) = \frac{\mathrm{i} t}{2\pi}$$ at its two simple poles. Let $t > 0$. The residue theorem yields $$\int \limits_{-r}^r g(t,x) \, \mathrm{d} x = 2 \pi \mathrm{i} \operatorname{Res}(g(t,\cdot),\mathrm{i}) - I_r(t) = - t - I_r(t) $$ for $r>1$, where $I_r(t)$ is the integral of $g(t,\cdot)$ along a semi-circle of radius $r$ in the upper half-plane. Since $g(t,\cdot)$ vanishes quadratically at infinity above the real axis, we have $\lim_{r \to \infty} I_r (t) = 0$ and $f(t) = - t = -\lvert t \rvert$ follows. For $t<0$ we can close the contour in the lower half-plane or just use the symmetry of $f$ to find $f(t) = t = -\lvert t \rvert$.
Reduction to the Dirichlet integral
For $t \in \mathbb{R} \setminus \{0\}$ we have \begin{align} f(t) &= \int \limits_{-\infty}^\infty g(t,x) \, \mathrm{d} x \stackrel{x \to - x}{=} \int \limits_{-\infty}^\infty g(t,-x) \, \mathrm{d} x = \int \limits_{-\infty}^\infty \frac{g(t,x) + g(t,-x)}{2} \, \mathrm{d} x \\ &= \int \limits_{-\infty}^\infty \frac{\cos(t x)-1}{\pi x^2} \, \mathrm{d} x \stackrel{\text{symmetry}}{=} - \frac{2}{\pi} \int \limits_0^\infty \frac{1 - \cos(\lvert t \rvert x)}{x^2} \, \mathrm{d} x \stackrel{u = \lvert t\rvert x}{=} - \lvert t \rvert \frac{2}{\pi} \int \limits_0^\infty \frac{1 - \cos(u)}{u^2} \, \mathrm{d} u \\ &= - \lvert t \rvert \frac{2}{\pi} \int \limits_0^\infty \frac{\sin(u)}{u} \, \mathrm{d} u = - \lvert t \rvert \, . \end{align}
