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How could it be proved that $$ \int_0^\infty J_0\left(\alpha\sqrt{x^2+z^2}\right)\ \cos{\beta x}\ \mathrm{d}x = \frac{\cos\left(z\sqrt{\alpha^2-\beta^2}\right)}{\sqrt{\alpha^2-\beta^2}} $$ for $0 < \beta < \alpha$ and $z > 0$ ? $J_0(x)$ is the zeroth order of Bessel function of the first kind.

I found this integral in Gradshteyn and Ryzhik's book 7th edition, section 6.677, the equation number 3. Any helps and hints will be appreciated!

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Note that for $r>0$ one has integral representation $$J_0(r)=\frac{1}{2\pi}\int_0^{2\pi}e^{ir\cos\phi}d\phi$$ Hence $$I=\int_0^{\infty}J_0\left(\alpha\sqrt{x^2+z^2}\right)\cos \beta x\,dx= \frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha\sqrt{x^2+z^2}\cos\phi}\cos\beta x \, d\phi.\tag{1}$$ On the other hand, $$\sqrt{x^2+z^2}\cos\phi=z\cos(\phi-\phi_0)+x\sin(\phi-\phi_0),$$ where $\tan\phi_0=-\frac{x}{z}$. Exchanging the order of integration in (1) and shifting $\phi$ by $\phi_0$, we arrive at $$I=\frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha(z\cos\phi+x\sin\phi)}\cos\beta x \, d\phi.$$ Finally, using that $\displaystyle\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}dx=\delta(\omega)$ we obtain $$I=\frac{1}{4}\int_0^{2\pi}e^{i\alpha z\cos\phi}\Bigl[\delta\left(\alpha\sin\phi+\beta\right)+\delta\left(\alpha\sin\phi-\beta\right)\Bigr]d\phi$$ It remains to use $\delta(f(x))=\sum\limits_{\text{zeros of }f}\frac{1}{|f'(x_k)|}\delta(x-x_k)$ and compute the two contributions coming from each of the two delta-functions.

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