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From here, our differentiated equation is <math> F'(a) = \int_{0}^{\infty}(-x*e^{-x^2}*sin{(a*x)}) dx</math>, which we can then integrate using integration by parts.
 
From here, our differentiated equation is <math> F'(a) = \int_{0}^{\infty}(-x*e^{-x^2}*sin{(a*x)}) dx</math>, which we can then integrate using integration by parts.
 
Doing so, however, would only get us:
 
Doing so, however, would only get us:
<center><math> \frac{sin{(a*x)}}{2e^{x^2}}\Big|_{0}^{\infty}</math></center>
+
<center><math> \frac{sin{(a*x)}}{2e^{x^2}}\Big|_{0}^{\infty} - \frac{a}{2}\int_{0}^(\infty)(e^{-x^2}*cos(a*x)) dx</math></center>
  
 
[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]
 
[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]

Revision as of 18:59, 27 November 2020

What is Feynman's Technique?

Feynman's Technique of integration utilizes parametrization and a mix of other different mathematical properties in order to integrate an integral that is can't be integrated through normal processes like u-substitution or integration by parts. It primarily focuses on setting a function equal to an integral, and then differentiating the function to get an integral that is easier to work with. A simple example would be an integral such as:

$ \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx $

As we can see, there isn't any particular place that we can use u-substitution or integration by parts to produce a solution easily, but Feynman shows us how we can parameterize the integral as a function, focusing on the cosine factor of the integrand. By writing the integral as a function, we can change the expression to:

$ F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx $ (where a = 2)

This allows us to extract an x from the cosine segment of the integrand by differentiating with respect to a, making the left portion of the integrand $ x*e^{-x^2} $, which is much easier to deal with than just $ e^{-x^2} $

From here, our differentiated equation is $ F'(a) = \int_{0}^{\infty}(-x*e^{-x^2}*sin{(a*x)}) dx $, which we can then integrate using integration by parts. Doing so, however, would only get us:

$ \frac{sin{(a*x)}}{2e^{x^2}}\Big|_{0}^{\infty} - \frac{a}{2}\int_{0}^(\infty)(e^{-x^2}*cos(a*x)) dx $

Back to Feynman Integrals

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood