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I am working on an elasticity problem which requires solving an integration with a rather complex kernel involving production of two Bessel's functions of the first kind (zeroth order) and a singularity. I have been struggling to solve it for almost one week but still have no clear clue. I will be really appreciate if someone could shed some light on it.

The integration I am working on writes as \begin{equation} K(s_j, s_i) = \int_{0}^{\infty} \frac{\lambda s_j J_{0}(\lambda s_i) J_{0}(\lambda s_j)}{\frac{e^{2h(2\pi - \lambda)}}{(1 + 2\lambda h + 2\lambda^2 h^2)} - (1-\theta) + \frac{4\lambda h}{(1 + 2\lambda h + 2\lambda^2 h^2)}}d\lambda \end{equation}

in which $s_i$ and $s_j$ are real values between $[0, a)$ and $s_i = s_j$ is an allowed case. $h$ is a rather big number which usually larger than 100. $\theta = (-1, 1)$ so that the integrand is singluar when \begin{equation} \frac{e^{2h(2\pi - \lambda)}}{(1 + 2\lambda h + 2\lambda^2 h^2)} - (1-\theta) + \frac{4\lambda h}{(1 + 2\lambda h + 2\lambda^2 h^2)} = 0 \end{equation}

Because $h$ is rather big, the singularity is always close to $\lambda = 2\pi$, which is within the range of integration. I do not have a clear strategy to deal with the singularity. And, even if this singularity could be well treated, the integration over $\lambda J(s_i\lambda)J(s_j\lambda)$ is still not an easy job. There seems to be an analytical form for it as \begin{equation} \int_{0}^{\infty} \lambda J_{0}(\lambda s_i) J_{0}(\lambda s_j)d\lambda = \frac{\delta(s_i - s_j)}{s_j} \end{equation}

in which $\delta$ is the Dirac function. This gives me another problem since it is not bounded... Am I thinking in the right track? Does this integral really sovlable? Thank you for your suggestions!

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  • $\begingroup$ If $\lambda_1$ is a positive root of $e^{2h(2\pi - \lambda)} - (1-\theta)(1 + 2\lambda h + 2\lambda^2 h^2) + 4\lambda h= 0$, then the integration will be singular at $\lambda=\lambda_1$, unless $J_0(\lambda_1 s_i)=0$ or $J_0(\lambda_1 s_j)=0$. $\endgroup$ – mike Jun 23 '14 at 22:01
  • $\begingroup$ Yes, that's the complexity embedded within the integration. $\lambda_1$ is always positive and always exists since the expression converges to $\theta - 1$, which is a negative value. $\endgroup$ – user159770 Jun 24 '14 at 14:26

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