Let $f: \mathbb{R}_{>0}\to \mathbb{R}$ be a function defined by $f(t)=\frac{\sin{t}}{t}+\frac{\sin(t)\cos(t)}{t}$. I need to find the Fourier transform of $f(t)$.
My approach: By definition of Fourier's transform, we know that $$\color{blue}{\boxed{\hat{f}(\xi)=\int_{\mathbb{R}}f(t)e^{i\xi t}\operatorname{dx}=\mathfrak{F}\{f(t) \}}}$$ and that the Fourier's transform is linear, it's to say: $\color{blue}{\boxed{\mathfrak{F}(f(t)+g(t))=\mathfrak{F}\{f(t)\}+\mathfrak{F}\{g(t)\}}}$. Then, we have that \begin{eqnarray*} \mathfrak{F}\{f(t)\}&=&\mathfrak{F}\left\{ \frac{\sin(t)}{t}+\frac{\sin(t)\cos(t)}{t} \right\}\\ &=&\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\}+\mathfrak{F}\left\{\frac{\sin(t)\cos(t)}{t} \right\} \end{eqnarray*} Now, we need calculate $$\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\} \quad \text{and} \quad \mathfrak{F}\left\{\frac{\sin(t)\cos(t)}{t} \right\}$$ I think, it's enough calculate the first transform, because then we can use after the convolution's theorem to invoke the transformation of $\frac{\sin(t)\cos(t)}{t}$ I know that the convolution's theorem say that: $\color{blue}{\boxed{\mathfrak{F}\{z(t)* \ell(t)\}=\mathfrak{F}\{z(t)\}\cdot\mathfrak{F}\{\ell(t)\}}}$ and taking $z(t)=\frac{\sin(t)}{t}$ and then $\ell(t)=\cos(t)$ we can follow the result.
Now, we can see that $$\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\}=\int_{\mathbb{R}}\frac{\sin(t)}{t}e^{it\xi}\operatorname{dx}:=\int_{\mathbb{R}}\operatorname{sinc}(t)e^{it\xi}\operatorname{dt}$$
but, $\operatorname{sinc}(t)\notin L^{1}(\mathbb{R})$, so I think we can't calculate the Fourier's transform using the definition maybe. So, how can I continue from here? Thanks to @reuns for pointing out that there is a problem that arises when applying the idea of convolution with the Fourier transform of the cosine. Then, how can I solve this problem?
First, we note that we can evaluate the Fourier Transform of $f(t)=\frac{\sin(t)}{t}$ function in a straightforward way by interpreting the Fourier Transform as the Cauchy Principal Value of the Riemann integral. Proceeding, we write
$$\begin{align} \mathscr{F}\{f\}(\xi)&=\text{PV}\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{i\xi t}\,dt\\\\ &=2\int_0^\infty \frac{\sin(t)\cos(\xi t)}{t}\,dt\\\\ &=\int_0^\infty \frac{\sin((\xi+1)t)-\sin((\xi-1)t)}{t}\,dt\\\\ &=\frac{\pi}{2}\left(\text{sgn}(\xi+1)-\text{sgn}(\xi-1)\right)\\\\ &=\begin{cases} \pi,& |\xi|<1\\\\ 0,&|\xi|>1\\\\ \pi/2,&|\xi|=1\tag1 \end{cases} \end{align}$$
Now, let $g(t)=f(t)\cos(t)=\frac{\sin(t)\cos(t)}{t}=\frac{\sin(2t)}{2t}$. Using $(1)$, the Fourier Transform of $g(t)$ is
$$\begin{align} \mathscr{F}\{g\}(\xi)&=\text{PV}\int_{-\infty}^\infty \frac{\sin(2t)}{2t}e^{i\xi t}\,dt\\\\ &=\frac12 \text{PV}\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{i\xi t/2}\,dt\\\\ &=\frac{\pi}{4}\left(\text{sgn}(\xi/2+1)-\text{sgn}(\xi/2-1)\right)\\\\ &=\begin{cases} \pi/2,& |\xi|<2\\\\ 0,&|\xi|>2\\\\ \pi/4,&|\xi|=2\tag2 \end{cases} \end{align}$$
If we wish to use the Convolution Theorem, then we need to appeal to the theory of distributions. Then, inasmuch as the Fourier Transform of $\cos(t)$ is $\frac12\left(\delta(\xi+1)+\delta(\xi-1)\right)$, where $\delta$ is the Dirac Delta, then we have
$$\begin{align} \mathscr{F}\{f\cos\}(\xi)&=\frac\pi4\int_{-\infty}^\infty (\delta(\xi-\tau-1)+\delta(\xi-\tau+1))\,\,\left(\text{sgn}(\tau+1)-\text{sgn}(\tau-1)\right)\,d\tau\\\\ &=\frac\pi4 \left(\text{sgn}(\xi+2)-\text{sgn}(\xi-2)\right)\\\\ &=\begin{cases} \pi/2,& |\xi|<2\\\\ 0,&|\xi|>2\\\\ \pi/4,&|\xi|=2 \end{cases} \end{align}$$
which agrees with the result in $(2)$!