Let me elaborate, 'Feynman's' trick ranks up in the top ten on most people's list, right behind contour integration, for best ways to evaluate definite integrals.

However, unlike contour integration , this method doesn't really seem to have a users manual on how and when to use the method. In addition I rarely see the trick used on this site, and even when I do, it's usually in conjunction with contour integration or some other trick, so it makes it hard to see exactly what property is being exploited.

Are there certain classes of integrals where using this method makes sense? In general, when is it best to use Feynman's trick?

Here are my own thoughts. $e^{p \cdot x}$ is an important kernel used in the Laplace transform. In addition, it's derivative with respect to $p$ adds an $x$ term to the integral increasing the likelihood that a trick like integration by parts would work. Thus, functions with Laplace transforms might be solvable using Feynman Integration.

  • 1
    $\begingroup$ See this PDF $\endgroup$ – user170231 Jun 27 '15 at 21:19
  • $\begingroup$ @user170231 I know what the technique is. I want the overarching class of definite integrals evaluable by this technique. The PDF only gives a list of examples, useful, but not what I wanted. $\endgroup$ – Zach466920 Jun 27 '15 at 21:46
  • $\begingroup$ The second page lists the conditions required in order to interchange the anti/differentiation operators. $\endgroup$ – user170231 Jun 27 '15 at 21:52
  • $\begingroup$ @user170231: I am not completely sure, but I think the question asks more about "fuzzy" criteria for recognizing when differentiation under the integral sign may be worthwile (i.e. simplify things), than formal criteria when it is justified. $\endgroup$ – PhoemueX Jun 27 '15 at 21:56
  • $\begingroup$ @user170231 Exactly what PeomueX said. $\endgroup$ – Zach466920 Jun 27 '15 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.