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:
∫∞0(e−x2∗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)=∫∞0(e−x2∗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−x2, which is much easier to deal with than just e−x2
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:
sin(a∗x)2ex2∣∣∞0−a2∫∞0(e−x2∗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 −a2F(a)
Thus, our result is F′(a)=−a2F(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 dFda=(−a2)F (source: blackpenredpen's explanation). With this, we can manipulate the equation to show dFF=−a2da
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
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:
∫∞0(e−x2∗cos(a∗x))dx=e−a24∗C1 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
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).
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):
F(a)=∫∞0(e−x2∗cos(a∗x))dx=e−a24∗π√2 Since our initial equation began with a value of a = 2, we can just plug in 2 for a into e−a24∗π√2
This gives us the answer to the integral, which is ∫∞0(e−x2∗cos(2x))dx=π√2e
