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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    $\begingroup$ Your v2 is now a duplicate of: this. There are plenty of questions here regarding this specific issue (i.e. KG propagator, Bessel functions etc.). $\endgroup$ Commented May 8 at 12:15
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    $\begingroup$ Either use a software package like Mathematica that can handle symbolic integration (but make sure you understand what it's doing under the hood since it can make different assumptions about things like branch cuts than you would make), or use a comprehensive integration table like Gradshteyn and Ryzhik. $\endgroup$
    – Andrew
    Commented May 8 at 12:17
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    $\begingroup$ What do you mean with find them?? I just searched for "bessel function qft". $\endgroup$ Commented May 8 at 12:18
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    $\begingroup$ What? There is a search bar?! It literally says: "Search on Physics..." You can search for words, users, comments, tag etc. There are also more advanced search in the SE data explorer tool (I don't have a link right now). $\endgroup$ Commented May 8 at 12:20
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    $\begingroup$ Sorry, but this is a problem totally on your side. I don't mean to be rude, but really this is a non-issue for the main site. It should be right next to your profile picture (left to it). If you don't see it, try a different browser, to start with. $\endgroup$ Commented May 8 at 12:22

1 Answer 1


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

  • $\begingroup$ So, would this be evident if I were to study the properties of Bessel functions? $\endgroup$ Commented May 8 at 12:31
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    $\begingroup$ Yes. It is one of the standard special functions and is usualy covered in math courses for physics. You should know that intergrals of exponentials of sin, cos, sinh, or cosh, are Bessel functions. $\endgroup$
    – mike stone
    Commented May 8 at 12:37
  • $\begingroup$ Thanks, I knew that I could write the integrand as an exponential of a trigonometric function, but was not aware of their connection to Bessel functions, and my CRC table of integrals does not list such integrals. I need to invest in Gradsteyn and Ryzhik. time to start studying Bessel functions. $\endgroup$ Commented May 8 at 12:54

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