In the fifth article of this series, BoJone derived some formulas that seemed unreasonable and did not explain their applications or origins. In fact, these were summarized while I was studying the following integral:
\int_{-\infty}^{+\infty} \frac{\cos x}{a^2+x^2}dx
Calculation of a Special Case
How should this integral be calculated? I didn’t have a good idea at first. Although it already provides a parameter a, directly differentiating with respect to a would make the integrand more complex without leading to an obviously integrable form. Therefore, I decided to first study a special case, namely when a=1:
\int_{-\infty}^{+\infty} \frac{\cos x}{1+x^2}dx
Since differentiating with respect to the original parameter did not simplify the expression, it indicates that Feynman’s integration method cannot be directly applied using that parameter. I decided to try changing the parameter to:
F(a)=\int_{-\infty}^{+\infty} \frac{\cos (ax)}{1+x^2}dx
Obviously, when a=0, the integral value is F(0)=\pi. According to the discussion in Feynman’s Integration Method (5), as a \to \infty, we have F(\infty)=0.
Let’s try differentiating: F'(a)=-\int_{-\infty}^{+\infty} \frac{x \sin (ax)}{1+x^2}dx
There is still no obviously integrable form. However, due to the periodicity of trigonometric differentiation, I was reminded of the method of solving integrals using differential equations mentioned in Feynman’s Integration Method (3). Thus, I differentiated once more:
\begin{aligned}F''(a)=&\,-\int_{-\infty}^{+\infty} \frac{x^2 \cos (ax)}{1+x^2}dx \\ =&\, \int_{-\infty}^{+\infty} \frac{\cos (ax)}{1+x^2}dx - \int_{-\infty}^{+\infty}\cos (ax)dx \\ =&\,F(a)-\int_{-\infty}^{+\infty}\cos (ax)dx\end{aligned}
What a surprise!! A second-order linear differential equation has emerged. According to the discussion in Feynman’s Integration Method (5), \int_{-\infty}^{+\infty}\cos (ax)dx=0. This means:
F''(a)=F(a)
This makes things much easier. We can directly write the general solution F(a)=C_1 e^a+C_2 e^{-a}. Given F(0)=\pi and F(\infty)=0, we immediately derive:
F(a)=\pi e^{-a}
At this point, we can obtain:
\begin{aligned}\int_{-\infty}^{+\infty} \frac{\cos x}{a^2+x^2}dx =&\, \frac{1}{a}\int_{-\infty}^{+\infty} \frac{\cos (a\times \frac{x}{a})}{1+(\frac{x}{a})^2}d\left(\frac{x}{a}\right) \\ =&\, \frac{1}{a}F(a)=\frac{\pi}{a}e^{-a}\end{aligned}
Reflections and Doubts
If you have derived this process yourself, I believe you would be as confused as I was by one thing: according to F'(a)=-\int_{-\infty}^{+\infty} \frac{x \sin (ax)}{1+x^2}dx, we should get F'(0)=0. However, the solution we found clearly does not follow this rule. Why is this?
At first, I was completely baffled. But then I thought, a=0 is a very special case. What happens if a \neq 0? I tried setting a to a very small specific value (such as 0.000001) and found that the result calculated from F'(a)=-\int_{-\infty}^{+\infty} \frac{x \sin (ax)}{1+x^2}dx is consistent with the result from F(a)=\pi e^{-a}. The final conclusion is: the integral \int_{-\infty}^{+\infty} \frac{x \sin (ax)}{1+x^2}dx is discontinuous at a=0!! In fact, one can think of it this way: while a approaches 0, x also approaches infinity. More importantly, even if \sin(ax) approaches 0, x \sin(ax) does not necessarily approach 0. In summary, it can be boiled down to one sentence:
In this integral, taking the limit with respect to a first and then x, versus taking the limit with respect to x first and then a, yields different results!
This warns us that the order of limits for functions of multiple variables cannot be swapped arbitrarily!
On the other hand, Feynman’s Integration Method (5) actually provides another way to solve: \int_{-\infty}^{+\infty} \frac{\cos x}{a^2+x^2}dx Which is to transform it into: \int_{-\infty}^{+\infty} \frac{e^{ix}}{a^2+x^2}dx And introduce a parameter to make it: F(b)=\int_{-\infty}^{+\infty} e^{-b(x^2+a^2)}\frac{e^{ix}}{a^2+x^2}dx Then differentiate with respect to b. This is also a straightforward approach, but the process requires using the conclusions from Feynman’s Integration Method (3). Interested readers might want to give it a try.
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