I need to solve an integral similar to the one in a book, Abramowitz and Stegun. Handbook of Mathematical Functions,P486, Eq. 11.4.29. However, I can't use infinity as the upper bound. Can someone give me some clues? Thanks in advance. The integral would be:

$$ \int_0^{d}e^{-a^2t^2} t^{v+1} J_v(bt)dt$$ or $$ \int_0^{d}e^{-a^2t^2} t J_0(bt)dt$$

  • $\begingroup$ If the upper limit is not infinity, then the integral does not possess a closed form when $d>0$ has a closed form, and vice-versa. $\endgroup$ – Lucian Feb 14 '15 at 6:09

Since $$ J_0(t) = \sum_{m\geq 0}\frac{(-1)^m}{m!^2}\left(\frac{t}{2}\right)^{2m} \tag{1}$$ and $$ b^{2m}\int_{0}^{d}t^{2m+1}e^{-a^2 t^2}\,dt =\frac{b^{2m}}{2 a^{2m+2}}\left(m!-\Gamma(m+1,a^2 d^2)\right)\tag{2}$$ we have: $$\begin{eqnarray*} \int_{0}^{d}e^{-a^2 t^2}t\,J_0(bt)\,dt &=& \sum_{m\geq 0}\frac{(-1)^m}{4^m m!}\cdot\frac{b^{2m}}{2a^{2m+2}}-\sum_{m\geq 0}\frac{(-1)^m}{4^m m!^2}\cdot\frac{b^{2m}}{a^{2m+2}}\Gamma(m+1,a^2 d^2)\\&=&\color{red}{\frac{1}{2a^2}e^{-\frac{b^2}{4a^2}}}-\color{blue}{\sum_{m\geq 0}\frac{(-1)^m}{4^m m!^2}\cdot\frac{b^{2m}}{a^{2m+2}}\Gamma(m+1,a^2 d^2)}\tag{3}\end{eqnarray*}$$ an expression (obviously) depending on the incomplete $\Gamma$ function.

  • $\begingroup$ OK, thanks everyone. I wonder why the book has only definite integration from zero to infinity. I'll use numerical integration then. $\endgroup$ – bjshhbyl Feb 14 '15 at 16:43

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