Solving dual integral equation I am trying to solve for $g(y)$ in the following set of dual integral equations:
$$\int_0^\infty  y \ g(y) \ J_0(xy) \ dy = 0\ \text{ for }\  0<x<1$$
$$\int_0^\infty g(y)\ J_0(yx) \ dy = x^0 \ \text{ for }\ 1<x<\infty $$
From other textbooks I know that for 
$$\int_0^\infty g(y)\ J_n(xy) \ dy=x^n\ \text{ for }\  0<x<1$$
$$\int_0^\infty y\ g(y)\ J_n(yx) \ dy =0\ \text{ for }\ 1<x<\infty $$
the solution will be given by
$$g(y)=\frac{2\Gamma(n+1)}{\sqrt{\pi}\Gamma(n+1/2)}j_n(y)$$
If anyone has any guidance as to how the first set of equations is solved, it would be very much appreciated
 A: For the system

\begin{align}
\begin{cases}
\int_0^\infty  y \ g(y) \ J_0(xy) \ dy &= f(x) &  0<x<1 \\
\int_0^\infty g(y)\ J_0(yx) \ dy &= g(x) & 1<x<\infty 
\end{cases}
\end{align}

the solution is
\begin{align}
g(x) &= \frac{2}{\pi} \, \int_{0}^{1} \left( \int_{0}^{t} \frac{u \, f(u) \, du}{\sqrt{u^2 - t^2}} \right) \, \sin(x t) \, dt - \frac{2}{\pi} \, \int_{1}^{\infty} \frac{d}{dt} \left( \int_{t}^{\infty} \frac{u \, g(u) \, du}{\sqrt{u^2 - t^2}} \right) \, \sin(x t) \, dt.
\end{align}
Integration by parts in the second integral leads to
\begin{align}
g(x) &= \frac{2}{\pi} \, \int_{0}^{1} \left( \int_{0}^{t} \frac{u \, f(u) \, du}{\sqrt{u^2 - t^2}} \right) \, \sin(x t) \, dt + \frac{2 \, x}{\pi} \, \int_{1}^{\infty} \left( \int_{t}^{\infty} \frac{u \, g(u) \, du}{\sqrt{u^2 - t^2}} \right) \, \cos(x t) \, dt \\
& \hspace{15mm} + \frac{2}{\pi} \, \sin(x) \, \int_{1}^{\infty} \frac{u \, g(u) \, du}{\sqrt{u^2 - 1}}.
\end{align}
If $f(x) = 0$ then the first integral is zero. If $g(x)$ is a constant then the integral involving $g(u)$ does not converge. 
This solution is a reduction based on equations (3.1) and (3.2) of B. N. Mandal, "A Note on Bessel function dual integral equation with weight function"
