Integration of Bessel Function of the first kind

Question

I have encountered a problem during my thesis study. I need to find $F(x)$ in terms of $G(y)$. My statement as follows: $$\int_{0}^\infty F(x)[Bx^3J_0(xy)+x^4J_1(xy)]dx=G(y)$$ where $B$ is a constant, $G(y)$ is an unknown function of $(n-1)$. degree polynomial, and $J_0$ and $J_1$ are the Bessel functions of the first kind. If $B=0$, F(x) could be found using Hankel transform as follows: $$x^3F(x)=\int_{0}^\infty yG(y) J_1(xy)\,dy$$

Question. However, if $B\neq0$ then how can I represent $F(x)$ in terms of $G(y)$? I want to obtain only $F$ an the left side of the equation, there can be anything at the right, derivative or integral of $G$.

Thanks.

Answer 1

With $\phi(x)=x^2F(x)$, your equation can be written as $G(x)=BI_0+I_1 $. As you notice, $$I_0=\mathcal{H}_0\phi(y)\triangleq\psi(y)$$ is a Hankel transform, and $$ I_1=\int_0^\infty\phi(x)xJ_1(xy)x\,dx=-\frac{d}{dy}\psi(y)$$ Then, the transform $\psi(y)$ is a solution of the ODE $$B\psi(y)-\psi'(y)=G(y)$$ which can be expressed as $$\psi(y)=Ce^{By}-e^{By}\int_0^y e^{-Bt}G(t)\,dt $$. From its integral representation $\psi'(0)=0$, $B\psi(0)=G(0)$ and thus $$\psi(y)=\frac{G(0)}{B}e^{By}-e^{By}\int_0^y e^{-Bt}G(t)\,dt $$ Now if this function admits an inverse (conditions on $G(y)$ and on $B$) you can express the result as a Hankel transform.