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Given $r,a,\lambda\in\mathbb{R}$, $r<a$, how can I find an approximate solution for the following definite integral?

$$ \int_0^\infty J_0 (\lambda r)J_1(\lambda a)\frac{1}{\sqrt{n+\lambda^2 }}\,d\lambda $$

$J_0,J_1$ are Bessel function of first kind, $0$th and $1$st order, respectively. $n$ is an imaginary parameter ($i\omega/\alpha$ where $\omega$ is angular frequency and $\alpha$ is thermal diffusivity) and cannot be considered small. $\lambda$ is the Fourier transform variable, $r$ is the radial coordinate and $a$ is a constant and can be considered small.

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    $\begingroup$ I would be glad to provide an answer, but I am paralyzed at the moment. It happens every time I see an integral with a completely useless extra parameter ($n$). $\endgroup$ – Jack D'Aurizio Aug 21 '15 at 2:07
  • $\begingroup$ In additon to @JackD'Aurizio remark, would u be so nice to tell us something about the other parameters? tend they to infinity or zero? how big are they compared to each other? $\endgroup$ – tired Aug 21 '15 at 9:45
  • $\begingroup$ For $n=0$ we have $\dfrac2\pi~E\bigg(\dfrac ra\bigg)$, see elliptic integrals for more information. $\endgroup$ – Lucian Aug 21 '15 at 14:44
  • $\begingroup$ $n$ is an imaginary parameter ($i\omega/\alpha$ where $\omega$ is angular frequency and $\alpha$ is thermal diffusivity) and cannot be considered small. $\lambda$ is the Fourier transform variable, $r$ is the radial coordinate and $a$ is a constant and can be considered small. $\endgroup$ – Ehsan Nasr Esfahani Aug 21 '15 at 17:41
  • $\begingroup$ Yes, but even so, it may be profitable to consider limits of $n$ both small and large. In that way, something like a two-point Pade approximant may be constructed. $\endgroup$ – Ron Gordon Aug 21 '15 at 19:40

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