| Line 16: | Line 16: | ||
Now, to proceed with the rest of the calculation, we need to express this equation in terms of differential fractions. By that, I mean writing it in the form <math> \frac{dF}{da} = (-\frac{a}{2})F</math> (source: blackpenredpen's explanation). With this, we can manipulate the equation to show <math>\frac{dF}{F} = -\frac{a}{2}da</math> | Now, to proceed with the rest of the calculation, we need to express this equation in terms of differential fractions. By that, I mean writing it in the form <math> \frac{dF}{da} = (-\frac{a}{2})F</math> (source: blackpenredpen's explanation). With this, we can manipulate the equation to show <math>\frac{dF}{F} = -\frac{a}{2}da</math> | ||
| − | After integrating both sides, we are left with <math>\ln{F} = -\frac{a^2}{4} + C</math>. By then considering ln as a logarithmic function of e, we can conclude that <math>F = e^{-\frac{a^2}{4}}*C_1</math> | + | After integrating both sides, we are left with <math>\ln{F} = -\frac{a^2}{4} + C</math>. By then considering ln as a logarithmic function of base "e", we can conclude that <math>F = e^{-\frac{a^2}{4}}*C_1</math> |
Finally, to solve for <math>C_1</math>, we substitute the original equation <math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> back into or new equation to get: | Finally, to solve for <math>C_1</math>, we substitute the original equation <math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> back into or new equation to get: | ||
Revision as of 19:24, 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) $
Now, to proceed with the rest of the calculation, we need to express this equation in terms of differential fractions. By that, I mean writing it in the form $ \frac{dF}{da} = (-\frac{a}{2})F $ (source: blackpenredpen's explanation). With this, we can manipulate the equation to show $ \frac{dF}{F} = -\frac{a}{2}da $
After integrating both sides, we are left with $ \ln{F} = -\frac{a^2}{4} + C $. By then considering ln as a logarithmic function of base "e", we can conclude that $ F = e^{-\frac{a^2}{4}}*C_1 $
Finally, to solve for $ C_1 $, we substitute the original equation $ F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx $ back into or new equation to get:
