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I want to know how to deduce $$ \int_0^\infty K_\nu(ax)I_\nu(bx)\cos cxdx=\frac{1}{2\sqrt{ab}}Q_{\nu-\frac12}(\frac{a^2+b^2+c^2}{2ab}) $$ My attempt:

I have evaluated $$ \int_0^\infty J_\nu(ax)J_\nu(bx)e^{-cx}dx=\frac{1}{2\sqrt{ab}}Q_{\nu-\frac12}(\frac{a^2+b^2+c^2}{2ab}) $$ Then I want to prove $$ \int_0^\infty K_\nu(ax)I_\nu(bx)\cos cxdx= \int_0^\infty J_\nu(ax)J_\nu(bx)e^{-cx}dx $$ I have tried Fourier transform, Mellin transform and series to prove the equation, but failed.

How to deduce the equation? $$ \int_0^\infty K_\nu(ax)I_\nu(bx)\cos cxdx=\frac{1}{2\sqrt{ab}}Q_{\nu-\frac12}(\frac{a^2+b^2+c^2}{2ab}) $$ or $$ \int_0^\infty K_\nu(ax)I_\nu(bx)\cos cxdx= \int_0^\infty J_\nu(ax)J_\nu(bx)e^{-cx}dx $$ Thank you for your time.

1 Answers1

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The correct expression is $$ \int_0^\infty J_\nu(ax)J_\nu(bx)e^{-cx}dx=\frac{1}{\pi\sqrt{ab}}Q_{\nu-\frac{1}{2}}\!\left(\frac{a^2+b^2+c^2}{2ab}\right) $$ under suitable conditions on $\nu$, $a$, $b$ and $c$ (see 6.612.3 in Gradshteyn and Ryzhik). Assume further that all the parameters are real. We deform the contour of integration from the positive real axis to the positive imaginary axis by an appeal to Cauchy's theorem, employ http://dlmf.nist.gov/10.27.E6, http://dlmf.nist.gov/10.27.E8 and http://dlmf.nist.gov/10.4.E3 to deduce \begin{align*} & \int_0^{ + \infty } {K_\nu (ax)I_\nu (bx)\cos (cx)dx} = \Re \int_0^{ + \infty } {K_\nu (ax)I_\nu (bx)e^{icx} dx} \\ & = \Re \int_0^{ + i\infty } {K_\nu (ax)I_\nu (bx)e^{icx} dx} = \Re \int_0^{ + \infty } {iK_\nu (ate^{\frac{\pi }{2}i} )I_\nu (bte^{\frac{\pi }{2}i} )e^{ - ct} dt} \\ & = \Re \int_0^{ + \infty } {ie^{\pi i\nu /2} K_\nu (ate^{\frac{\pi }{2}i} )J_\nu (bt)e^{ - ct} dt} = \frac{\pi }{2}\Re\int_0^{ + \infty } {H^{(2)}_\nu (at)J_\nu (bt)e^{ - ct} dt} \\ & = \frac{\pi }{2}\int_0^{ + \infty } {J_\nu (at)J_\nu (bt)e^{ - ct} dt} =\frac{1}{2\sqrt{ab}}Q_{\nu-\frac{1}{2}}\!\left(\frac{a^2+b^2+c^2}{2ab}\right). \end{align*} We may omit the condition that the parameters are real via analytic continuation. For the precise conditions, see 6.672.4 in Gradshteyn and Ryzhik.

Gary
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