The integral equations \[ F(x) = \frac{d}{{dx}}\int_0^x {J_0 [2\sqrt {k(x - t)} ]f(t)dt} \] and \[ G(x) = \frac{d}{{dx}}\int_x^\infty {J_0 [2\sqrt {k(t - x)} ]g(t)dt} \] are usually considered separately, and their solutions \[ f(x) = \frac{d}{{dx}}\int_0^x {I_0 [2\sqrt {k(x - t)} ]F(t)dt} \] and \[ g(x) = \frac{d} {{dx}}\int_x^\infty {I_0 [2\sqrt {k(t - x)} ]G(t)dt} \] respectively are regarded unique. These solutions in general are not square integrable. We prove that under certain conditions either one of these two integral equations gives the square integrable solution of the other. Moreover, \[ \frac{d}{{dx}}\int_{ - \infty }^x {J_0 [2\sqrt {k(x - t)} ]f(t)dt} = 0 \qquad {\text{and}}\qquad \frac{d}{{dx}}\int_x^\infty {J_0 [2\sqrt {k(t - x)} ]g(t)} = 0,\quad - \infty < x < \infty , \] have nontrivial solutions. Therefore the assumption regarding the uniqueness of the solution of the second integral equation is not valid unless some additional conditions are specified.It is well known that the homogene...