A friend of mine was the TA for a graduate level Math course for Physicists. And an exercise in that course was to solve integrals using Contour Integration. Just for fun, I decided to mess with him by trying to solve all the contour integral problems in the prescribed textbook for the course [Arfken and Weber’s ‘Mathematical methods for Physicists,7th edition” exercise (11.8)] using anything BUT contour integration.
You can solve a lot of them them exclusively by using Feynman’s trick. ( If you would like to know about what the trick is – here is an introductory post) The following are my solutions:
*I forgot how to solve these 4 problems without using Contour Integration. But I will update them when I remember how to do them. If you would like, you can take these to be challenge problems and if you solve them before I do send an email to 153armstrong(at)gmail.com and I will link the solution to your page. Cheers!
Parametric Integration is an Integration technique that was popularized by Richard Feynman but was known since Leibinz’s times. But this technique rarely gets discussed beyond a niche set of problems mostly in graduate school in the context of Contour Integration.
A while ago, having become obsessed with this technique I wrote this note on applying it to Laplace transform problems and it is now public for everyone to take a look.
Parametric integration is one such technique that once you are made
aware of it, you will never for the love of god forget it. It goes by many names : ‘Differentiation under the Integral sign’, ‘Feynman’s famous trick’ , ‘Parametric Integration’ and so on.
Now this integral might seem familiar to you if you have taken a calculus course before and to evaluate it is rather simple as well.
Knowing this you can do lots of crazy stuff. Lets differentiate this
expression wrt to the parameter in the integral – s (Hence the name
parametric integration ). i.e
Look at that, by simple differentiation we have obtained the expression
for another integral. How cool is that! It gets even better.
Lets differentiate it once more:
If you keep on differentiating the expression n times, one gets this :
Now substituting the value of s to be 1, we obtain the following
integral expression for the factorial. This is known as the gamma
There are lots of ways to derive the above expression for the gamma
function, but parametric integration is in my opinion the most subtle
way to arrive at it. 😀