Applications to Physics & Mathematics

When used in different integrals, Feynman's integral can simplify mathematicians' and students' lives. We can use this technique in solving arduous definite and improper definite integrals. To better apply this technique, physicists use this trick to solve problems in quantum physics. They tweak the equations or functions and introduce ideas from complex numbers to simplify their functions. To give an example of how this is done, let's look at the following example.

When given a definite integral such as,

$ \int_{0}^{\pi} e^{cos(x)}cos(sin(x)) dx $

Using what we learnt from feynman's technique, we can modify this as a function of:

$ T(b) = \int_{0}^{\pi} e^{bcos(x)}cos(bsin(x)) dx $

Mathematicians and Physicists would tweak this function in a simpler way so that they would be able to use feynman's technique. Introducing Euler's formula from complex numbers, will help solving this problem a lot easier: $ e^{ix} = cos(x) + isin(x) $ Before we start computing, we transform the integrand slitghly with whatever information or ideas we know about:

$ T(b) = \int_{0}^{\pi} e^{bcos(x)}cos(bsin(x)) dx = \frac{1}{2}\int_{-\pi}^{\pi} e^{bcos(x)}cos(bsin(x)) dx $ $ = \frac{1}{2} \int_{0}^{2\pi} e^{bcos(x)}cos(bsin(x)) dx $ From the equation, $ e^{ix} = cos(x) + isin(x) $, we can get $ e^{isinx} = cos(sinx) + isin(sinx) $. But with the problem we only want to deal with the real part of the complex number, and hence we will just use $ Re(e^{isinx}) $ which is $ cos(sin x) $. Therefore, we can write our function as $ T(b) = \frac{1}{2} \int_{0}^{2\pi} e^{bcos(x)}Re(e^{ibsinx}) dx $

A question may arise about how we deal when we are just using the real part of a complex number with a real number. To answer that, real numbers and complex numbers form a Banach space. Banach space theory is an arduous topic to learn about in mathematics as it involves theories like Group theory and others, but to answer the question in simple words, we see that both parts of the integrands are real, and hence we can consider the whole function as:

$ T(b) = Re (\frac{1}{2} \int_{0}^{2\pi} e^{bcos(x)}(e^{ibsinx}) dx) = Re (\frac{1}{2} \int_{0}^{2\pi} e^{be^{ix}}dx) $

With the problem now posed like this, we can continue to use Feynman's technique:

$ T'(b) = \frac{1}{2} \frac{d}{db} \int_{0}^{2\pi} e^{be^{ix}}dx = \frac{1}{2} \int_{0}^{2\pi} \frac{\partial}{\partial b} e^{be^{ix}}dx $ $ = \frac{1}{2} \int_{0}^{2\pi} ibe^{be^{ix}}e^{ie} dx = \frac{1}{2} e^{be^{ix}} \Big|_{0}^{2\pi} $ $ = 0 $

Since our derivative is 0, we know that $ T(b) $ is a constant w.r.t. b. Therefore, we can conclude that

$ T(b) = T(0) = \int_{0}^{\pi} dx = \pi $

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