Computing the Integral $\int r^2 \text{J}_0(\alpha r) \text{I}_1(\beta r)\text{d}r$

Question

I encountered the following integral in a physical problem $$I=\int r^2 \text{J}_0(\alpha r) \text{I}_1(\beta r)\text{d}r$$ where $\text{J}_0$ is the Bessel function of first kind of order $0$ and $\text{I}_1$ is the modified Bessel function of order $1$. Also, $\alpha$ and $\beta$ are arbitrary real numbers.

It seems that MAPLE and WOLFRAM are not able to find the primitive. However, I think that there should be a tidy one in terms of Bessel functions.

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My Thought

My instinct in integration tells me to use integration by parts and the recursive relations for Bessel functions. But I couldn't get any where yet.

Answer 1

First note that

$$ r\text{I}_1(\beta r)=\frac{\partial}{\partial\beta} \text{I}_0\left(\beta r\right).$$

Then we can write \begin{align} \mathcal{I}(t):=&\int_0^t r^2 \text{J}_0\left(\alpha r\right)\text{I}_1\left(\beta r\right) dr\\ =&\frac{\partial}{\partial\beta}\int_0^t r\text{J}_0\left(\alpha r\right)\text{I}_0\left(\beta r\right)dr\\ =& \frac{\partial}{\partial\beta}\left[\frac{t\left(\alpha \text{J}_1(\alpha t)\text{I}_0(\beta t)+\beta \text{J}_0(\alpha t)\text{I}_1(\beta t)\right)}{\alpha^2+\beta^2}\right]. \end{align} The last line follows from the formula 1.11.5.2 in Vol. II of Prudnikov-Brychkov-Marychev (and can of course be verified by direct differentiation).

Answer 2

According to the Start Wearing Purple's answer, I just took the computations and wanted to put the final answer here

$$\eqalign{ & I = {1 \over {{\alpha ^2} + {\beta ^2}}}\,\,\,\,\left( {\alpha {\text{I}_1}\left( {\beta r} \right){\text{J}_1}\left( {\alpha r} \right) + \beta {\text{I}_0}\left( {\beta r} \right){\text{J}_0}\left( {\alpha r} \right)} \right){r^2} \cr & \,\,\, - {{2\beta } \over {{{\left( {{\alpha ^2} + {\beta ^2}} \right)}^2}}}\left( {\alpha {\text{I}_0}\left( {\beta r} \right){\text{J}_1}\left( {\alpha r} \right) + \beta {\text{I}_1}\left( {\beta r} \right){\text{J}_0}\left( {\alpha r} \right)} \right)r \cr} $$

Hope that it may be useful for future readers. There may still be other heuristic ways to compute this integral so do not hesitate to write a new answer with a different approach. :)