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Is there any solution for this integral? $$ \int_a^b xe^{-\alpha x^2}J_n(\beta x) dx\,. $$

I looked up in all books i could find and only found this:

$$ \int_0^{\infty} xe^{-\alpha x^2}J_n(\beta x) dx=\frac{\sqrt{\pi}\beta}{8\alpha^{\frac{3}{2}}} \exp \left( -\frac{\beta^2}{8\alpha}\right)\left[ I_{\frac{1}{2} n-\frac{1}{2} } \left(\frac{\beta^2}{8\alpha} \right)- I_{\frac{1}{2} n+\frac{1}{2} } \left( \frac{\beta^2}{8\alpha}\right)\right] \,, $$ for $\Re[\alpha] >0, and \ \ \ \Re[n]>-2$

Should I use a Gaussian distribution to approximately transform this integral into a definite one?

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Well, I guess the answer is "no" if you look for a closed form solution.
But the series representation can be obtained. One can use the expansion of the Bessel function: $$ J_n(\beta x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+n+1)} {\left(\frac{\beta x}{2}\right)}^{2m+n} $$

$$\int_a^b xe^{-\alpha x^2}J_n(\beta x)\mathrm dx=\sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+n+1)} {\left(\frac{\beta}{2}\right)}^{2m+n}\int_a^b e^{-\alpha x^2}x^{2m+n+1}\mathrm dx$$ Setting (for the sake of simplicity) $\kappa=2m+n+1$ and keeping in mind that (for positive $n$) $\kappa>1$ $$\int_a^b e^{-\alpha x^2}x^{\kappa}\mathrm dx=\frac{1}{2} \alpha ^{\frac{1-\kappa }{2}} \left(\Gamma \left(\frac{\kappa +1}{2},a^2 \alpha \right)-\Gamma \left(\frac{\kappa +1}{2},b^2 \alpha \right)\right)$$ So the initial integral (in case of positive $n$, $a$ and $b>a$) will look like: $$\int_a^b xe^{-\alpha x^2}J_n(\beta x)\mathrm dx\!=\\ \frac{1}{2\sqrt{\alpha}}\!\!\sum_{m=0}^\infty \!\frac{(-1)^m}{m! \, \Gamma(m+n+1)}\!{\left(\!\!\frac{\beta}{2 \sqrt{\alpha}}\!\right)\!}^{2m+n}\left(\!\Gamma \!\left(\!m\!+\!\frac{n}{2}\!+\!1,\left(a\sqrt{\alpha}\right)^2\!\right)\!\!-\!\Gamma \!\left(\!m+\!\frac{n}{2}\!+1,\left(b\sqrt{\alpha}\right)^2 \!\right)\!\right) $$ where $\Gamma(\cdot,\cdot)$ - is the incomplete Gamma function.
But I guess that the obtained series cannot be simplified. Surely it can be written in a more fancy way (through hypergeometric functions for example) with a shorter notation, but that will not change the result. :)

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