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Note: this is not a duplicate: I am not interested in the issue of the contour, but in methods of integration.

I am desirous to integrate the following: $$\int_m^{\infty}{\rho e^{-\rho r}\over\sqrt{\rho^2-m^2}}\; d\rho.$$ This integral occurs in the context of the authors' proof that the free Klein Gordon field theory is causal, i.e. it is the transition amplitude for space-like intervals. In an attempt to integrate, I have done the following: $$\begin{align}\int{\rho e^{-\rho r}\over\sqrt{\rho^2-m^2}}\;d\rho &=\int 2e^{-\rho r}{d\over d\rho}\bigg(\sqrt{\rho^2-m^2}\bigg)\;d\rho\cr &=2e^{-\rho r}\sqrt{\rho^2-m^2}+\int2re^{-\rho r}\sqrt{\rho^2-m^2}\;d\rho,\end{align}$$ where an integration by parts has been performed. Here is where I am stuck, I can find no integrable in my table that would allow me to go further. According to research on the topic, the integration can be accomplished in terms of Bessel functions, however, how does one get from the integral all the way to Bessel functions?

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There is a natural subsititution $\rho = m\cosh t$ that will turn the integral into the standard integral for the Bessel $K$ function: $$ \int_0^\infty \cosh \alpha t e^{-mr \cosh t} dt= K_\alpha(m\rho). $$

mike stone
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