Suppose that the autocorrelation function of a random signal $X(t)$ is given by $$R_{XX}(\tau)=e^{-|\tau|}$$
and $X(t)$ was passed to the input of an impulse response function $h$ defined as $h(t)=e^{-t}$ if $t>0$, and $0$ otherwise. Now I want to calculate the autocorrelation function of the output $Y(t)=h(t) * x(t)$.
I tried calculating $Y(t)$ first: $$Y(t)=\int_{-\infty}^{\infty} h(\tau)x(t-\tau)d\tau=\int_0^{\infty}e^{-\tau}x(t-\tau)d\tau$$ but got stuck from here.
The second approach I've tried was implying Fourier Transform on $R_{XX}(\tau)$ and $h(t)$:
$$|X(w)|^2 =\int_{-\infty}^{\infty} e^{-|t|}e^{-jwt}dt=\frac{2}{w^2+1}, H(w)=\int_0^{\infty} e^{-(jw+1)t}dt=\frac{1}{jw+1}$$ and $|Y(w)|^2=|H(w)|^2|X(w)|^2=\frac{1}{w^2+1}\cdot\frac{2}{w^2+1}=\frac{2}{(w^2+1)^2}$, but I couldn't calculate the inverse Fourier transformation of $\frac{2}{(w^2+1)^2}$.
I think there should be an easy idea to solve this problem. How can I solve this? Any help is appreciated.
Let $F(\omega)=\frac2{(\omega^2+1)^2}$. The inverse Fourier Transform, $f$, of $F$ is given by
$$\begin{align} f(t)&=\mathscr{F}^{-1}\{F\}(t)\\\\ &=\frac1{2\pi}\int_{-\infty}^\infty \frac2{(\omega^2+1)^2}e^{i\omega t}\,dt\\\\ \end{align}$$
Note that we can differentiate $f$ twice by differentiating under the integral to find
$$\begin{align}f''(t)&=-\frac1{2\pi}\int_{-\infty}^\infty \frac{2\omega^2}{(\omega^2+1)^2}e^{i\omega t}\,d\omega\\\\ &=\underbrace{-\frac1{2\pi}\int_{-\infty}^\infty \frac{2}{\omega^2+1}e^{i\omega t}\,d\omega}_{=-e^{-|t|}}+\underbrace{\frac1{2\pi}\int_{-\infty}^\infty \frac2{(\omega^2+1)^2}e^{i\omega t}\,d\omega}_{=f(t)} \end{align}$$
Therefore, we find that $f''(t)-f(t)=-e^{-|t|}$ with $f(0)=1/2$ and $f'(0)=0$. Now, solving the ODE subject to the initial conditions yields
$$f(t)=\frac12 (1+|t|)e^{-|t|}$$
Alternatively, we can use contour integration to evaluate the Fourier Transform. Note that there are poles at $\pm i$. Therefore, we have for $t>0$
$$\begin{align} f(t) &= i\text{Res}\left(\frac{2e^{i\omega t}}{(\omega^2+1)^2}, \omega=i\right)\\\\ &=i \lim_{\omega \to i}\frac{d}{d\omega}\left(\frac{2e^{i\omega t}}{(\omega+i)^2}\right)\\\\ &=i\left(\frac{i2t e^{-t}}{-4}-\frac{4e^{-t}}{-8i}\right)\\\\ &=\frac12(t+1)e^{-t} \end{align}$$
Inasmuch as $f(t)$ is even, we have
$$f(t)=\frac12 (1+|t|)e^{-|t|}$$
as expected!