This exercise gave me nightmares this night. I have $$ x(t)=\sin(t^2)e^{-2|t-2|} $$ to Fourier transform.
First I though about solving the integral. (should I divide the signal in $2$, first for $t-2<0$ then for $t-2\ge 0$ and then study then separately?) I then express the sin in exponentials with Euler and write up the Fourier integral (edit: which is also wrong). Well I'm not able to solve this integral, so I guess there is a smarter way to go about it, any hint?
My convention for the Fourier transform is $$\widehat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x\xi}\mathrm{d}x,\qquad \xi\in\mathbb{R} \tag{1}$$ You can adjust to your own convention using the scaling properties of the Fourier transform.
As Chester noted in the comments, you have a product of two functions $\sin t^{2}$ and $e^{-2\left|t-2\right|}$. However, the convolution theorem $\widehat{fg}=\widehat{f}\ast\widehat{g}$ only applies for suitable functions $f$ and $g$ or a distribution $f$ and a suitable function $g$. In any case, the Fourier transform of $\sin t^{2}$ only exists in the sense of distributions: $$\langle{\sin(\cdot)^{2},\widehat{\phi}}\rangle=\int_{\mathbb{R}}\sin(t^{2})\widehat{\phi}(t)\mathrm{d}t,\qquad\forall\phi\in\mathcal{S}(\mathbb{R})$$ As I don't know how familiar you are with distribution theory, I am going to try to avoid referring to it.
First, observe that by writing
$$\sin(t^{2})=\dfrac{e^{it^{2}}-e^{-it^{2}}}{2i}, \tag{2}$$ it suffices to replace $\sin t^{2}$ by $e^{\pm it^{2}}$ and then use linearity.
By the dominated convergence theorem, we have
$$\lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}e^{-\delta x^{2}}e^{\pm ix^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\int_{\mathbb{R}}e^{-ix^{2}}e^{-2\left|x-2\right|}\mathrm{d}x$$ For any $\delta>0$, the LHS is the Fourier transform of two $L^{1}$ functions $e^{-\delta x^{2}}e^{-ix^{2}}$ and $e^{-2\left|x-2\right|}$, which is given by the convolution of their Fourier transforms. Alternatively, we have a product of two square-integrable functions, so we can use Plancherel's theorem.
Lemma 1. If $f_{\delta}(x)=e^{-(\delta +i\sigma)x^{2}}$, for $\delta>0$ and $\sigma\in\mathbb{R}$, then $$\widehat{f_{\delta}}(\xi)=\int_{\mathbb{R}}e^{-(\delta+i\sigma)x^{2}}e^{-2\pi i x\xi}\mathrm{d}x=\left(\dfrac{\pi}{\delta+i\sigma}\right)^{1/2}e^{-(\pi\xi^{2})/(\delta+i\sigma)}, \tag{3}$$ where we take the principal branch of $\arg(z)$.
This result is typically proven using Cauchy's theorem from complex analysis, and a proof can be found in a typical Fourier analysis book. Since we will need it later below, we note that
$$\lim_{\delta\rightarrow 0^{+}}\widehat{f_{\delta}}(\xi)=\left(\dfrac{\pi}{\left|\sigma\right|}\right)^{1/2}e^{-\text{sgn}(\sigma)i\pi/4}e^{i\pi\xi^{2}/\sigma},\qquad\forall\xi\in\mathbb{R} \tag{4}$$
Lemma 2. If $g(x)=e^{-2\left|x-2\right|}$, then $$\widehat{g}(\xi)=\dfrac{e^{-4\pi i\xi}}{1+\pi^{2}\xi^{2}} \tag{5}$$
Proof. It suffices by the translation and scaling properties to compute the Fourier transform of the function $e^{-\left|t\right|}$l. \begin{align*} \int_{\mathbb{R}}e^{-\left|x\right|}e^{-2\pi i x\xi}\mathrm{d}x=\int_{0}^{\infty}e^{-x}2\cos(2\pi x\xi)\mathrm{d}x \end{align*} Integrating by parts twice, we obtain that the RHS is equal to $$-(2\pi^{2}\xi^{2})^{-1}\int_{0}^{\infty}e^{-x}\cos(2\pi x\xi)\mathrm{d}x,$$ which with some algebra yields $$\int_{\mathbb{R}}e^{-\left|x\right|}e^{-2\pi i x\xi}\mathrm{d}x=\dfrac{2}{1+4\pi^{2}\xi^{2}}$$ $\Box$
Applying Lemma 1 with $\sigma=\pm 1$ and Lemma 2, we obtain that
\begin{align*} \lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}e^{-(\delta\pm i) x^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\lim_{\delta\rightarrow 0^{+}}\int_{\mathbb{R}}\left(\dfrac{\pi}{\delta\pm i}\right)^{1/2}e^{-(\pi y^{2})/(\delta\pm i)}\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \end{align*} The integrand on the RHS above is dominated by an $L^{1}$ function for all $\delta>0$, so by dominated convergence, we conclude that
\begin{align*} \int_{\mathbb{R}}e^{\mp ix^{2}}e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x=\sqrt{\pi}e^{\mp i\pi/4}\int_{\mathbb{R}}e^{\pm i(\pi y^{2})}\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \tag{6} \end{align*}
Whence, \begin{align*} \int_{\mathbb{R}}\sin(x^{2})e^{-2\left|x-2\right|}e^{-2\pi i x\xi}\mathrm{d}x&=\dfrac{\sqrt{\pi}}{2i}\int_{\mathbb{R}}\left[e^{-i(\pi y^{2}-\pi/4)}-e^{i(\pi y^{2}-\pi/4)}\right]\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y\\ &=-\sqrt{\pi}\int_{\mathbb{R}}\sin(\pi y^{2}-\pi/4)\dfrac{e^{-4\pi i(\xi-y)}}{1+\pi^{2}(\xi-y)^{2}}\mathrm{d}y \tag{7} \end{align*}
At the moment, I don't know how to obtain a closed form expression for the integral on the right. Perhaps another user could chime in on this point.
