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I was searching in the Prudnikov (vol. 2) how to solve an integral and I finally found it. However, I didn't recognized a function that appears in the answer.


$$ \int_0^x x^{\nu+1} e^{a x^2} J_{\nu}(b x) dx = \frac{b^\nu e^{a x^2}}{(2 i a)^{\nu+1}} \left[U_{\nu+1} (2iax^2, bx) + i U_{\nu+2} (2iax^2, bx)\right],~~~[Re{\nu} > -1] $$

I have searched what function is "$U_{\nu}(\cdot, \cdot)$" but I didn't found. Initially I believed that it was a Chebyshev polynomial, but it takes two parameters, not only one...

Anyone have an idea? I need to know in order to try to simplify the final expression.

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$U_n(x)$ denotes sometimes the $n$-th Chebyshev polynomial of the 2nd kind. It is related to the representation theory of $SU(2)$. Does your integral originate from representation theory, say the group $GL(2,C)$? –  Marc Palm Feb 6 at 19:11
Have you checked integral tabels, see e.g. here: mathoverflow.net/questions/99027/… –  Marc Palm Feb 6 at 19:13
@MarcPalm, the answer is no. My integral appeared when I was trying to obtain a cumulative distribution function (i.e., integrating a probability density function). Gradshteyn and Ryzhik don't have the integral, but I found it in Prudnikov (the integral is shown in the question). The only change I have made in my integral is converting the $I_\nu(\cdot)$ in its $J_\nu(\cdot)$ representation –  Rafael Feb 9 at 5:04

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