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

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 ?

• – Harry Peter Jul 21 '14 at 16:42

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}

• I think the equation 2 is missing a denominator term of $\sqrt t$, also I am not quite sure how to calculate this corrected equation (2). – Comic Book Guy Jul 20 '14 at 0:02
• Can u please also explain how you got equation 1 from the first equation you had, should n't there be a $\sin$ term ? – Comic Book Guy Jul 20 '14 at 0:22
• @Hardy: Equation (1) is does read $\cos(a y^{2}) + i \sin(a y^{2})$. It is more general if the notion that two integrals can be calculated at once. It also makes it possible to condense some integrals into a more manageable form, such as is the case here. Also, from an earlier comment, for $t = y^{2}$, notice that $dt = 2 y \ dy$ and the integral contains the element $y \ dy$ as stated. – Leucippus Jul 20 '14 at 0:29
• Thank you for correcting me about the sqrt term. – Comic Book Guy Jul 20 '14 at 0:41