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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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I know it's a little late to reply, but just in case someone else needs this integral, the book Tables of Integrals, Series, and Products, 7th Ed. (Eq. 1, section 6.631, pg. 755), gives this integral as, \begin{align} \int_0^{\infty} x^{\mu} e^{-\alpha x^2} J_{\nu}\left(\beta x \right) dx &= \frac{ \beta^{\nu} \Gamma\left(\frac{1}{2}\nu + \frac{1}{2}\mu + \frac{1}{2}\right) } { 2^{\nu+1} \alpha^{\frac{1}{2}(\nu+\mu+1)}\Gamma(\nu+1) } {}_1F_1\left( \frac{\nu+\mu+1}{2}; \nu+1; -\frac{\beta^2}{4\alpha} \right). \end{align}

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