Does $\int { y\cosh \left(\beta y^2\right)}J_0\left(\gamma y^2 \right) dy$ have a closed form

Question

I am trying to solve the following indefinite integral

$$F_Y(y) = \int {y\cosh \left(\beta y^2\right)}J_0\left(\gamma y^2 \right) dy$$

Where $J_0$ is the Bessel function of the first kind.

I tried to expand the $J_0$ using it's Taylor Series expansion like J.M showed here but the expression became much too unwieldy afterwards.

Is it possible to express this integral in a closed form (preferably, using elementary functions, Bessel functions, integers and basic constants) maybe with hyper geometric function ?

Answer 1

Some notes to consider:

1. Let $\beta \rightarrow ia$ to obtain the form \begin{align} F(a, \gamma) = \int y \cos(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy \end{align} or, more generally, \begin{align}\tag{1} F(a, \gamma) = \int y e^{ia y^{2}} \ J_{0}(\gamma y^{2}) \ dy. \end{align}
2. Let $t = y^{2}$ for which (1) becomes \begin{align}\tag{2} F(a, \gamma) = \frac{1}{2} \int e^{i at} \ J_{0}(\gamma t) \ dt. \end{align}

After the calculation of (2) take the real part of the value obtained and then let $a \rightarrow - i \beta$ to obtain the desired result.

Example: If the limits are $(0, \infty)$ then \begin{align} \int_{0}^{\infty} e^{-iat} \ J_{0}(\gamma t) \ dt = \frac{-i}{\sqrt{a^{2} - \gamma^{2}}}. \end{align} This leads to \begin{align} \int_{0}^{\infty} y \ \cos(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= 0 \\ \int_{0}^{\infty} y \ \sin(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= \frac{1}{\sqrt{a^{2} - \gamma^{2}}}. \end{align} In the desired form of what the question is seeking the results are \begin{align} \int_{0}^{\infty} y \ \cosh(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= 0 \\ \int_{0}^{\infty} y \ \sinh(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy &= \frac{-1}{\sqrt{a^{2} + \gamma^{2}}}. \end{align}