Proof of Gradshteyn and Ryzhik (6.637.1)

Question

I want to prove that $$ F := \int_0^\infty \dfrac{e^{-a \sqrt{x^2 + b^2}} }{\sqrt{x^2 + b^2}} J_\nu (cx) \, dx = I_{\nu/2} \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} - a \right] \right) K_{\nu/2} \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} + a \right] \right), $$ which is (6.637.1) from Gradshteyn and Ryzhik.

The first thing that I have tried is to make the substitution $x = b \sinh \zeta$, so $$ F = \int_0^\infty e^{-ab \cosh \zeta} J_\nu (bc \sinh \zeta) \, d\zeta. $$ Now I introduce $$ J_\nu (r) = \dfrac{(r/2)^\nu}{\sqrt{\pi} \, \Gamma(\nu + 1/2)} \int_0^\pi \cos(r \cos\theta) (\sin \theta) ^{2\nu} \, d\theta, $$ so that \begin{equation} F = \int_0^\pi (\sin \theta) ^{2\nu} \int_0^\infty \dfrac{(bc/2 \sinh \zeta)^\nu}{\sqrt{\pi} \, \Gamma(\nu + 1/2)} e^{-ab \cosh \zeta} \cos(bc \sinh \zeta \cos\theta) \, d\zeta\, d\theta. \label{1}\tag{1} \end{equation} I now want to use the fact that $$ I_\nu (r) = \dfrac{(r/2)^\nu}{\sqrt{\pi} \, \Gamma(\nu + 1/2)} \int_0^\pi \cosh(r \cos\theta) (\sin \theta) ^{2\nu} \, d\theta, $$ and $$ K_\nu (r) = \dfrac{\sqrt{\pi} \, (r/2)^\nu}{\Gamma(\nu + 1/2)} \int_0^\infty e^{-r \cosh\zeta} (\sinh \zeta) ^{2\nu} \, d\zeta, $$ and probably also the facts that $$ \dfrac{b^2c^2}{4} = \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} - a \right] \right) \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} + a \right] \right) $$ and $$ -ab = \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} - a \right] \right) - \left(\dfrac{b}{2} \left[ \sqrt{a^2 + c^2} + a \right] \right) $$ to put \eqref{1} into the desired form, but I am not sure how to treat the crossed term involving $\sinh\zeta \cos\theta$. It seems like I might also have to use contour integration to turn the hyperbolic cosine into the regular cosine. Can anyone help me fill in the details?

Answer 1

We start with $$F = \int_0^\infty e^{-ab\cosh \zeta} J_\nu (bc \sinh \zeta) d\zeta$$ Use [A1] to transform the exponent: $$F = \frac{2ab}{\sqrt{\pi}} \int_0^\infty \cosh \zeta \left[\int_0^\infty \exp \left(-t^2 a^2 b^2 \cosh^2 \zeta - \frac{1}{4t^2}\right)dt\right] J_\nu (bc \sinh \zeta) d\zeta$$ $$=\frac{2ab}{\sqrt{\pi}} \int_0^\infty \exp \left(-t^2a^2b^2 - \frac{1}{4t^2}\right) \left[\int_0^\infty e^{-t^2 a^2 b^2 \sinh^2 \zeta} J_\nu (bc \sinh \zeta) \cosh \zeta d\zeta\right]dt$$ $$=\frac{2ab}{\sqrt{\pi}} \int_0^\infty \exp \left(-t^2a^2b^2 - \frac{1}{4t^2}\right) \left[\int_0^\infty e^{-t^2 a^2 b^2 \zeta^2} J_\nu (bc \zeta) d\zeta\right] dt$$ Use [A2] to compute the integral with respect to $\zeta$: $$F=\int_0^\infty \exp \left(-t^2a^2b^2 - \frac{1}{4t^2}\right) \exp\left(-\frac{c^2}{8t^2 a^2}\right) I_{\nu/2} \left(\frac{c^2}{8t^2 a^2}\right) \frac{dt}{t}$$ $$=\frac12 \int_0^\infty \exp \left(-ta^2b^2 - \frac{1}{4t}-\frac{c^2}{8t a^2}\right) I_{\nu/2} \left(\frac{c^2}{8t a^2}\right) \frac{dt}{t}$$ $$=\frac12 \int_0^\infty \exp \left(-\frac{t}{2} - \frac{a^2b^2}{2t}-\frac{b^2c^2}{4t}\right) I_{\nu/2} \left(\frac{b^2c^2}{4t}\right) \frac{dt}{t}$$ Let $\xi = \frac{b}{2} \left(\sqrt{a^2+c^2}+a\right), z = \frac{b}{2} \left(\sqrt{a^2+c^2}-a\right)$, then $$F=\frac12 \int_0^\infty \exp \left(-\frac{t}{2} - \frac{\xi^2 + z^2}{2t}\right) I_{\nu/2} \left(\frac{\xi z}{t}\right) \frac{dt}{t}$$ Using [A3] we can get $$I_{\nu/2} \left(\frac{\xi z}{t}\right) = t \exp \left( \frac{\xi^2+z^2}{2t}\right) \int_0^\infty J_{\nu/2} (\xi v) J_{\nu/2} (z v) e^{-tv^2/2} v dv$$ Therefore $$F=\frac12 \int_0^\infty J_{\nu/2} (\xi v) J_{\nu/2} (z v) v dv \int_0^\infty \exp \left(-\frac{t}{2}\left(1+v^2\right)\right) dv dt$$ $$=\int_0^\infty J_{\nu/2} (\xi v) J_{\nu/2} (z v) \frac{v dv}{1+v^2}$$ Finally, we use [A4] to get $$F = I_{\nu/2}(z) K_{\nu/2}(\xi) = I_{\nu/2} \left(\frac{b}{2} \left[ \sqrt{a^2 + c^2} - a \right] \right) K_{\nu/2} \left(\frac{b}{2} \left[ \sqrt{a^2 + c^2} + a \right] \right)$$

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Appendix

[A1]. The Poisson integral representation (derivation): $$e^{-x} = \frac{2x}{\sqrt{\pi}} \int_0^\infty \exp \left(-t^2x^2 - \frac{1}{4t^2}\right) dt$$ [A2]. Gradshteyn and Ryzhik 6.618.1 (Watson^∗ §13.3) $$\int_0^\infty e^{-\alpha x^2} J_\nu (bx) dx = \frac{\sqrt{\pi}}{2\sqrt{\alpha}} \exp \left(-\frac{b^2}{8\alpha}\right) I_{\nu/2} \left(\frac{b^2}{8\alpha}\right),$$ $$\operatorname{Re}\alpha > 0, \quad b>0, \quad \operatorname{Re} \nu > -1$$ [A3]. Gradshteyn and Ryzhik 6.633.2 (Watson^∗ §13.31) $$\int_0^\infty e^{-g^2 x^2} J_p(ax) J_p (\beta x) x dx = \frac{1}{2g^2} \exp \left(-\frac{a^2+\beta^2}{4g^2}\right) I_p \left( \frac{a\beta}{2g^2}\right),$$ $$\operatorname{Re}p > -1, \quad |\operatorname{arg} g| < \pi/4, \quad a>0,\quad \beta>0$$ [A4]. Gradshteyn and Ryzhik 6.541.1 (see also the following post for the idea of derivation) $$\int_0^\infty x J_\nu (ax) J_\nu (bx) \frac{dx}{x^2+c^2} = I_\nu (bc) K_\nu (ac),$$ $$ \quad a > b, \quad \operatorname{Re} c > 0, \quad \operatorname{Re} \nu > -1$$ ^∗Watson, G.N. "A Treatise on the Theory of Bessel Functions", Cambridge University Press, 1922.