Probability density of sum of random variables from inversion formula

Question

Let $X, Y$ be two independent random variables on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ with densities $f_X (x) = \frac{1}{2} e^{-|x|}$ (double exponential) and $f_Y (x) = \frac{1}{\pi(1+x^2)}$ (Cauchy).

Let $Z=X+Y$. Thus, we know that the density of $Z$ is the convolution of the densities:
$$f_Z(x) = \int_{-\infty}^{\infty} f_X(x-y)f_Y(y) dy = \int_{-\infty}^{\infty} \frac{e^{-|x-y|}}{2 \pi (1+y^2)} dy \tag{1}$$

My problem is: if I try to get the same density using the characteristic functions of $X$ and $Y$ then I get a different result. Indeed:
$$\varphi_X(t) = \frac{1}{1+t^2}$$
$$\varphi_Y(t) = e^{-|t|}$$
$$\varphi_Z(t) = \varphi_X(t) \varphi_Y(t) = \frac{e^{-|t|}}{1+t^2}$$
We have that $\int_{\mathbb{R}} |\varphi_Z(t)| < \infty$, so we can use the inversion formula i.e.:
$$f_Z(x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{-ixy} \varphi_Z(y) dy = \frac{1}{2\pi} \int_{\mathbb{R}} e^{-ixy} \frac{e^{-|y|}}{1+y^2} dy \tag{2}$$

What went wrong? Why is $(1)$ different from $(2)$?

Answer 1

I found out that there was no problem: the two integrals are the same. In particular, $$g(x)=\frac{1}{2 \pi} \int_{\mathbb{R}} e^{-ixy}\frac{e^{-|y|}}{1+y^2}dy = \frac{1}{2 \pi} \int_{\mathbb{R}} \cos(xy) \frac{e^{-|y|}}{1+y^2}dy$$ by symmetry, and it happens to be equal to $$f(x)=\frac{1}{2 \pi} \int_{\mathbb{R}} \frac{e^{-|x-y|}}{1+y^2}dy$$ The two integrals are not trivial to compute, but Mathematica helped me: evaluating the integrals numerically with the code

f[x] := NIntegrate[Exp[-Abs[x - y]]/(1 + y^2), {y, -Infinity, Infinity}];
g[x] := NIntegrate[Cos[x*y]*Exp[-Abs[y]]/(1 + y^2), {y, -Infinity, Infinity}];
Plot[f[x], {x, -20, 20}, PlotStyle -> Green, PlotRange -> {{-20, 20}, {0, 1.5}}]
Plot[g[x], {x, -20, 20}, PlotRange -> {{-20, 20}, {0, 1.5}}]

We get the plot of $f(x)$:

and the plot of $g(x)$:

Also evaluating at random values we get the same value, for instance: $$f(1)=g(1)=0.958821$$ up to Mathematica precision.

This is not what I expected but I suppose it is an original way to check that these two integrals are the same!