One can try to do the following derivations.
$$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx$$
You can simplify the power of the exponent:
$$-\left(\frac{x}{\alpha}+\frac{1}{2w^2}(x-\hat{x})^2\right)=-\left(\frac{(x-\mu)^2}{2w^2}+\gamma \right),$$
where $\gamma=\left(\frac{\hat{x}}{\alpha}-\frac{w^2}{2\alpha^2}\right)$, $\mu=\hat{x}-\frac{w^2}{\alpha^2}$. So:
$$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{-\frac{(x-\mu)^2}{2w^2}} \, I_0\left(\frac{x}{\beta}\right)\,dx$$
Then you can change the variable of integration $y=x-\mu$:
$$\mathrm{Int}=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, (y+\mu) \, e^{-\frac{y^2}{2w^2}} \, I_0\left(\frac{y+\mu}{\beta}\right)\,dx$$
and make use of the Neumann’s addition theorem:
$$\mathop{I_{{\nu}}}\nolimits\!\left(u\pm v\right)=\sum _{{k=-\infty}}^{\infty}(\pm 1)^{k}\mathop{I_{{\nu+k}}}\nolimits\!\left(u\right)\mathop{I_{{k}}}\nolimits\!\left(v\right)$$
setting $\nu=0$, assuming that $y,\mu,\beta\in\mathbb{R}$ and using connection formulas one will get:
$$\mathop{I_{0}}\nolimits\!\left(\frac{y+\mu}{\beta}\right)=\sum _{{k=-\infty}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)=\mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{0}}}\nolimits\!\left(\frac{\mu}{\beta}\right)+2\sum _{{k=1}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)$$
$$\mathrm{Int}=\!
\frac{\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)e^{-\gamma}}{\sqrt{2\pi w^2}}\!\int_{-\infty}^{+\infty} \, (y+\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\!
+\!\frac{2 e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty}\! (y+\mu) \, e^{-\frac{y^2}{2w^2}} \! \sum _{k=1}^{\infty}\!\mathop{I_k}\!\!\left(\frac{y}{\beta}\right)\!\!\mathop{I_k}\nolimits\!\left(\frac{\mu}{\beta}\right) \,dx$$
Or reorganising terms:
$\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)$
$$\mathrm{Int}=\!
\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\!\left(\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)\int_{-\infty}^{+\infty} \, (y\!+\!\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\!
+\!2\!\sum _{k=1}^{\infty}\!\mathop{I_k}\!\left(\frac{\mu}{\beta}\right)\!\int_{-\infty}^{+\infty}\! (y\!+\!\mu) \, e^{-\frac{y^2}{2w^2}} \! \!\mathop{I_k}\!\left(\frac{y}{\beta}\right)\!\!\,dx \right)$$
Let's set
$$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$
$$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$
Here one can play with the evenness/unevenness of the functions under integrals (as I did at first), or just use CAS and get:
$$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\sqrt{\frac{\pi \alpha ^2}{8 \beta ^2}}\! e^{\frac{\alpha ^2}{4 \beta ^2}} \left(\!\!2 \beta \left(\!\!(-1)^k+1\!\right)\! \mu I_{\frac{k}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+\alpha ^2 \left(\!1-(-1)^k\!\right)\!\! \left(\!\!I_{\frac{k-1}{2}}\!\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+I_{\frac{k+1}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)\!\!\right)\!\!\right)$$
$$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\! \sqrt{2 \pi } \alpha \mu e^{\frac{\alpha ^2}{4 \beta ^2}} I_0\left(\frac{\alpha ^2}{4 \beta ^2}\right)$$
So the original integral can be represented in the following way:
$$\mathrm{Int}(\mu,\alpha,\beta)\!=\! \frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\left(\mathrm{Int}_0(\mu,\alpha,\beta)+
2\!\sum _{k=1}^{\infty}\mathop{I_k}\left(\frac{\mu}{\beta}\right)
\mathrm{Int}_k(\mu,\alpha,\beta)
\right)$$
And I am not shure that there is any chance to simplify it futher (meaning finding the sum of the series).
Prudnikov-Brychkov-Marychev book (Vol. 2) contains the following formula:
$$ \int_{0}^{\infty}x^{\alpha-1}e^{-px^2}I_{\nu}(cx)dx=2^{-\nu-1}c^{\nu}p^{-\frac{\alpha+\nu}{2}}\frac{\Gamma(\frac{\alpha+\nu}{2})}{\Gamma(\nu+1)}\, _1F_1\left(\frac{\alpha+\nu}{2},\nu+1,\frac{c^2}{4p}\right).$$
Here $\mathrm{Re}\,p>0$, $\mathrm{Re}(\alpha+\nu)>0$, $-\pi<\arg c<\pi$, and $_1F_1$ denotes confluent hypergeometric function. For $\nu=0$ this gives
$$ \int_{0}^{\infty}x^{\alpha-1}e^{-px^2}I_{0}(cx)dx=\frac12p^{-\frac{\alpha}{2}}\Gamma(\alpha/2)\; _1F_1\left(\frac{\alpha}{2},1,\frac{c^2}{4p}\right).$$
Further, for integer $\alpha$ (evens are better than odds) $_1F_1$ simplifies. For example,
$$ _1F_1(1/2,1,z)=e^{z/2}I_0(z/2),\qquad _1F_1(1,1,z)=e^z.$$
Similar answers for greater integer values of $\alpha$ can be obtained using recursion formulas for $_1F_1$ w.r.t. parameters.
Best Answer
Lets start from the definition of the modified Bessel Function. Recall
$$I_{0}(z)=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{z^{2n}}{n!n!}.$$ So our integral is
$$\int_{0}^{\infty}e^{-\frac{1}{2}\left(x^{2}+b^{2}\right)}x\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{b^{2n}x^{2n}}{n!n!}dx.$$ Switching the order of sumation and integration we get
$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx.$$ Now, let $u=\frac{1}{2}x^{2}$ and $du=xdx$, to see that $$\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=2^{n}\int_{0}^{\infty}e^{-u}u^{n}du=2^{n}\Gamma(n+1)=2^{n}n!.$$ Hence
$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\left(2^nn!\right)$$
$$=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{\left(\frac{b^{2}}{2}\right)^{n}}{n!}=e^{-\frac{1}{2}b^{2}}e^{\frac{1}{2}b^{2}}=1$$
as desired.
Notice that the exact same argument shows $$\int_0^\infty e^{\left(\frac{1}{2}b^2-\frac{1}{2}x^2\right)}J_0(bx)xdx=1$$ where $J_0(x)$ is the Bessel Function.
Hope that helps,