(Created page with "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...")
 
 
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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:
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= Worked Solution 1=
  
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∫∞0(e−x2∗cos(2x))dx
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A simple example would be an integral such as:
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<center><math> \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx</math></center>
 
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:
 
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:
 +
<center><math> F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx</math> (where a = 2)</center>
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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 <math>x*e^{-x^2}</math>, which is much easier to deal with than just <math>e^{-x^2}</math>
  
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F(a)=∫∞0(e−x2∗cos(a∗x))dx (where a = 2)
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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.
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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−x2, which is much easier to deal with than just e−x2
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Doing so, however, would only get us:
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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>
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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>
  
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From here, our differentiated equation is F′(a)=∫∞0(−x∗e−x2∗sin(a∗x))dx, which we can then integrate using integration by parts. Doing so, however, would only get us:
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Thus, our result is <math>F'(a) = -\frac{a}{2}F(a)</math>
  
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sin(a∗x)2ex2∣∣∞0−a2∫∞0(e−x2∗cos(a∗x))dx
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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>
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With this, we can see that the left side of the subtraction operation evaluates to 0, while the right side is just −a2F(a)
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Thus, our result is F′(a)=−a2F(a)
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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>
  
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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 dFda=(−a2)F (source: blackpenredpen's explanation). With this, we can manipulate the equation to show dFF=−a2da
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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:
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<center><math>\int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*C_1</math></center>
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Since the function works for any number "a", we can choose the variable to be set as 0, giving us <math>\int_{0}^{\infty}(e^{-x^2}) dx = e^{-\frac{a^2}{4}}*C_1</math>
  
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After integrating both sides, we are left with lnF=−a24+C. By then considering ln as a logarithmic function of base "e", we can conclude that F=e−a24∗C1
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Although we haven't covered it, this is the Gaussian integral, which, for the sake of this explanation, we will just take the value as <math>\frac{\sqrt{\pi}}{2}</math>(if you would like to know how this is calculated, there are probably solutions on YouTube or Google).
  
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Finally, to solve for C1, we substitute the original equation F(a)=∫∞0(e−x2∗cos(a∗x))dx back into or new equation to get:
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Because of this, we now know that the value of <math>C_1</math> is the Gaussian integral, which gives us our last equation which we set equal to our original F(a):
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<center><math>F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2}</math></center>
 +
Since our initial equation began with a value of a = 2, we can just plug in 2 for a into <math>e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2}</math>
  
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∫∞0(e−x2∗cos(a∗x))dx=e−a24∗C1
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This gives us the answer to the integral, which is <math>\int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx = \frac{\sqrt{\pi}}{2e}</math>
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Since the function works for any number "a", we can choose the variable to be set as 0, giving us ∫∞0(e−x2)dx=e−a24∗C1
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Although we haven't covered it, this is the Gaussian integral, which, for the sake of this explanation, we will just take the value as π√2(if you would like to know how this is calculated, there are probably solutions on YouTube or Google).
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[[ Walther MA271 Fall2020 topic14 | Back to Feynman Integrals]]
  
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Because of this, we now know that the value of C1 is the Gaussian integral, which gives us our last equation which we set equal to our original F(a):
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[[Category:MA271Fall2020Walther]]
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F(a)=∫∞0(e−x2∗cos(a∗x))dx=e−a24∗π√2
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Since our initial equation began with a value of a = 2, we can just plug in 2 for a into e−a24∗π√2
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This gives us the answer to the integral, which is ∫∞0(e−x2∗cos(2x))dx=π√2e
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Latest revision as of 00:34, 2 December 2020

Worked Solution 1

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 $

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:

$ \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*C_1 $

Since the function works for any number "a", we can choose the variable to be set as 0, giving us $ \int_{0}^{\infty}(e^{-x^2}) dx = e^{-\frac{a^2}{4}}*C_1 $

Although we haven't covered it, this is the Gaussian integral, which, for the sake of this explanation, we will just take the value as $ \frac{\sqrt{\pi}}{2} $(if you would like to know how this is calculated, there are probably solutions on YouTube or Google).

Because of this, we now know that the value of $ C_1 $ is the Gaussian integral, which gives us our last equation which we set equal to our original F(a):

$ F(a) = \int_{0}^{\infty}(e^{-x^2}*cos(a*x)) dx = e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2} $

Since our initial equation began with a value of a = 2, we can just plug in 2 for a into $ e^{-\frac{a^2}{4}}*\frac{\sqrt{\pi}}{2} $

This gives us the answer to the integral, which is $ \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx = \frac{\sqrt{\pi}}{2e} $

Back to Feynman Integrals

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