We can consider the general type of integral, means
$\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx$
Case 1. If $b=0$, the function identically 0, so the integral converges and equals to 0.
Remark. The function is an odd function of $b$ (we consider $b$ as a variable), so we can only consider cases of $b>0$ in the following.
Case 2. If $a<0$, the integral divergent and didn't exist.
Case 3. If $a>0$, we can calculate as following:
$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx &=\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}(\int_{0}^{b}\cos(xy)dy)dx\\ &=\int_{0}^{b}(\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\cos(xy)dx)dy \end{align*}$
$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\cos(xy)dx &=\int_{-\infty}^{\infty}e^{-ax^{2}}\cos((x+x_{0})y)dx\\ &=\cos(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx -\sin(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\sin(xy)dx\\ &=\cos(x_{0}y)\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx \end{align*}$
We denote $\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx$ by $F(y)$, then
$\begin{align*} F^{\prime}(y)&=-\int_{-\infty}^{\infty}xe^{-ax^{2}}\sin(xy)dx =\frac{1}{2a}\int_{-\infty}^{\infty}\sin(xy)de^{-ax^{2}}\\ &=-\frac{y}{2a}\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(xy)dx =-\frac{y}{2a}F(y) \end{align*}$
By calculation, we can obtain that $F(0)=\int_{-\infty}^{\infty}e^{-ax^{2}}dx =2\int_{0}^{\infty}e^{-ax^{2}}dx =\frac{1}{\sqrt{a}}\Gamma(\frac{1}{2}) =\frac{\sqrt{\pi}}{\sqrt{a}}$.
Then solve the ordinary differential equation with initial value, we can get:
$F(y)=F(0)e^{-\frac{y^{2}}{4a}}=\frac{\sqrt{\pi}}{\sqrt{a}}e^{-\frac{y^{2}}{4a}}$
So
$\begin{align*} \int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx &=\int_{0}^{b}\cos(x_{0}y)F(y)dy\\ &=\frac{\sqrt{\pi}}{\sqrt{a}}\int_{0}^{b}e^{-\frac{y^{2}}{4a}}\cos(x_{0}y)dy\\ &=\sqrt{\pi}\int_{0}^{\frac{b}{\sqrt{a}}}e^{-\frac{t^{2}}{4}}\cos(\sqrt{a}x_{0}t)dt \end{align*}$
Case 4. If $a=0$, then the integral becomes $\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx$, by the criterion of sigular integral, we know that the integral converges.
In particular,
$\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx =\lim_{a\rightarrow0^{+}}\int_{-\infty}^{\infty}e^{-a(x-x_{0})^{2}}\frac{\sin(bx)}{x}dx$
Set $a\rightarrow0^{+}$ in case 3, we can obtain that $\int_{-\infty}^{\infty}\frac{\sin(bx)}{x}dx =\sqrt{\pi}\int_{0}^{\infty}e^{-\frac{t^{2}}{4}}dt =\sqrt{\pi}\Gamma(\frac{1}{2})=\pi$