Any suggestions how to solve this or how to find an approximate solution \begin{equation} \int_0^a\int_0^\infty J_0 (\lambda r)J_1(\lambda a)\frac{1}{\sqrt{n+\lambda^2 }}d\lambda dr \end{equation}

(J0,J1 Bessel function of first kind, 0th and 1st order, respectively) Thanks, Ehsan

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    $\begingroup$ Do you have a source or motivation for this integral, and do you expect it to have a "simple" closed form? Also, what have you tried yourself? $\endgroup$ – mickep Jul 28 '15 at 20:09
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    $\begingroup$ Hi, given an periodic flux applied on a circular region (radius a) of a half space, the result of this integral is the averaged temperature over the circular area in frequency domain. In fact the results of this integral is a transfer function, relating the power to the temperature and I expect to have a simple closed form solution due to the symmetry. I tried different methods and I can solve the integral over r, but the integral over $\lambda$ seems difficult. Any suggestion how to solve it? $\endgroup$ – Ehsan Nasr Esfahani Jul 28 '15 at 20:41
  • $\begingroup$ Any approximate solution for this integral? $\endgroup$ – Ehsan Nasr Esfahani Aug 20 '15 at 17:29
  • $\begingroup$ Wolfram alpha exceeds its standard computational time when given this. Have you evaluated it numerically? $\endgroup$ – Neal Aug 20 '15 at 18:20

For $n=0$ Mathematica seems to think the integral is equal to $$ I= \frac{a}{4\pi}\left(4 + 8G-i\pi^2 \right) $$ where $G$ is the Catalan constant. However, numerically for $a=1$ it seems there is a branch problem giving the $-i\pi^2$. So it seems more like $$ I= \frac{a}{\pi}\left(1 + 2G\right) $$ it solved the internal integral over $\lambda$ to be $$ \frac{2}{a \pi r}\left(E\left(\frac{a^2}{r^2}\right)+(a-r)(a+r)K\left(\frac{a^2}{r^2}\right) \right) $$ for complete elliptic $E$ and $K$ functions, beware that the Mathematica notation is with elliptic modulus $k^2=a^2/r^2$ here.

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  • $\begingroup$ I didn't realise I had $n=0$ set in program, will work on general case now. $\endgroup$ – Benedict W. J. Irwin Jun 27 '17 at 11:14

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