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<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> | <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> | ||
With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just <math>-\frac{a}{2}F(a)</math> | With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just <math>-\frac{a}{2}F(a)</math> | ||
| + | |||
Thus, our result is <math>F'(a) = -\frac{a}{2}F(a)</math> | Thus, our result is <math>F'(a) = -\frac{a}{2}F(a)</math> | ||
[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]] | [[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]] | ||
Revision as of 20:07, 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:
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:
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:
With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just $ -\frac{a}{2}F(a) $
Thus, our result is $ F'(a) = -\frac{a}{2}F(a) $
