Let $X$ be a random variable such that its characteristic function is
$$\varphi_X(t)=e^{-|t|} \qquad \forall t \in \mathbb R$$
Calculate its distribution function.
My attempt at a solution:
The integral $$\int_{- \infty} ^{\infty} |\varphi_X(t)| \,dt$$ it's finite. So $X$ is absolutely continuous and, fixed $x \in \mathbb R$ , it's distribution function calculated in $x$ is equal to
\begin{align} f_X(x) &= \frac{1} {2\pi} \int_{- \infty} ^{\infty} e^{-itx} \varphi_X(t)\,dt \\ &= \frac{1} {\pi} \int_{0} ^{\infty} e^{-itx} e^{-t} \,dt \\ &= \frac{1} {\pi} \int_{0} ^{\infty} \cos(tx) e^{-t} \,dt - i \frac{1} {\pi} \int_{0} ^{\infty} \sin(tx) e^{-t} \,dt \\ &= \frac{1} {\pi} \frac {1} {1+x^2} - i \frac{1} {\pi} \frac {x} {1+x^2} \\ \end{align}
But the solution of the book is:
$$ f_X(x) = \frac{1} {\pi} \frac {1} {1+x^2} $$
So my doubt is: why the imaginary part disappears?
There is a mistake in line 2 of your proof. The integral from $-\infty$ to $+\infty$ is twice the integral form $0$ to $\infty$ for even functions. For odd functions the integral is actually $0$. So integral of $\sin (tx) \phi_X (t)$ is $0$.