I am having trouble with computing the convolution of $f(x)=\sin x/x$ and:
\begin{equation} g(x)=\begin{cases} 1-|x|,& -1 \leq x \leq 1 \\ 0, & x \notin [-1,1] \end{cases} \end{equation}
I have found until now that:
\begin{equation} (f*g)(x)=\int_{-\infty}^{\infty}f(x-y)g(y)dy=\int_{-\infty}^{\infty}g(x-y)f(y)dy=\\ \int_{-1}^{1}\frac{\sin(x-y)}{x-y}dy+\int_{-1}^{0}y\frac{\sin(x-y)}{x-y}dy+\int_{0}^{1}y\frac{\sin(x-y)}{x-y}dy \end{equation}
and for
\begin{equation} I_1=\int_{-1}^{1}\frac{\sin(x-y)}{x-y}dy \end{equation}
I can see that for $z=x-y$ I have: \begin{equation} I_1=\int_{x-1}^{x+1}\frac{\sin z}{z}dz=\frac{1}{2i}\hat{f}(-1)-\frac{1}{2i}\hat{f}(1) \end{equation}
where, $f(z)=1/z$ and $\hat{f}(k)=-i\sqrt{\frac{\pi}{2}}\text{sgn}(k)$ where I define the FT of $f$ as: \begin{equation} \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x)dx \end{equation}
The problem is that I know that the Fourier transform of $f(x)=1/x$ is the one I mentioned above but for the integral limits to be equal to $\pm \infty$. How can I express the same FT but for the integral limits to be equal to $[x-1,x+1]$ as above so that I can use the $\hat{f}(\pm 1)$ of the FT of $f(x)=1/x$?
Also, how can I compute the rest of the integrals remaining at the convolution?
Thank you!