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| − | + | A simple example would be an integral such as: | |
<center><math> \int_{0}^{\infty}(e^{-x^2}*cos(2x)) dx</math></center> | <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: | ||
Revision as of 22:54, 1 December 2020
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:
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):
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} $
