I found this integral in Gradshteyn-Ryzhik's book, $$ \int_a^\infty\ J_0\left(b\sqrt{x^2-a^2}\right)\ \sin(cx) \mathrm{d}x = \frac{\cos\left(a\sqrt{c^2-b^2}\right)}{\sqrt{c^2-b^2}}; ~~\mathrm{for~0<b<c}, $$ or can be written as $$ \int_0^\infty\ J_0\left(by\right)\ \sin\left(c\sqrt{y^2+a^2}\right) \frac{y}{\sqrt{y^2+a^2}} \mathrm{d}y = \frac{\cos\left(a\sqrt{c^2-b^2}\right)}{\sqrt{c^2-b^2}}; ~~\mathrm{for~0<b<c}. $$
Can I extend the integral into imaginary value of $a$? In other words, is it correct to write the integral below?
$$ \int_0^\infty\ J_0\left(by\right)\ \sin\left(c\sqrt{y^2-a^2}\right) \frac{y}{\sqrt{y^2-a^2}} \mathrm{d}y = \frac{\cosh\left(a\sqrt{c^2-b^2}\right)}{\sqrt{c^2-b^2}}; ~~\mathrm{for~0<b<c}. $$