DUTIS — Feynman’s Ingenious Integration Technique

I’ve recently found myself quite enamored by integrals and integration techniques. Amidst a plethora of unconventional techniques, I found myself vying with a particularly interesting one: DUTIS. Or, as it’s otherwise known, Richard Feynman’s Trick. What is DUTIS? Well, it’s short for Differentiating Under The Integral Sign. And, although it carries with it immense utility and power, it is hardly taught formally. So, here I am. Writing about it.

Before we begin though, here’s a little history: although this trick is credited to Feynman, he wasn’t the one to develop it. As written in his autobiography ‘Surely You’re Joking, Mr. Feynman’, DUTIS is a technique he adopted from Advanced Calculus, a textbook by F. Woods. Who, I imagine, picked it up from someone else. In fact, the use of this technique dates as far back as Leibniz and is formally called the Leibniz Integral Rule.

Now, we’re ready to dive into the good stuff. While there is a much stronger version of this technique that involves variable limits, we will leave that for the future, simply because of the incredibly heavy machinery it demands. However, a more elementary version of the theorem with constant limits of integration (satisfying most necessities) reads the following:

Note: Keep in mind, that both integrands need to be continuous with respect to both variables for the theorem to hold.

Why is this true? Well, the proofs are a little bit long but the interested reader can find beautiful proofs for both the basic and the full versions of this theorem here. But intuitively, this should make sense to you because of the fundamental theorems of calculus.

While the theorem seems all high and mighty written out, it’s of no use if it can’t be applied. So, to understand its application, we’re going to get our hands dirty.

DUTIS in Action

Our first example is a seemingly straightforward integral:

But I defy you to try and evaluate this using conventional methods. After you’ve given it your best shot, continue reading!

Our first task, in order to use DUTIS, is to introduce a dummy variable:

Now, we are in a position to apply Feynman’s trick. By differentiating the LHS and taking the partial derivative with respect to t of the integrand on the RHS, we get the following:

We suddenly see that our work is trivialized. Just one trick has made the integral incomparably easier. Now, all we need is to solve a simple separable ODE. Notice that I(2) = I. So, by the fundamental theorems of calculus, our works is reduced to evaluating the following integral:

And this is trivially integrable. Our result, therefore, is ln(3). This example is just one of many that demonstrate the vast capabilities of this ingenious trick. Another application is given below.

Say Bye-Bye to Contours (Only Sometimes, Though)

Contour integration is often tiring and tedious. One has to develop a contour, find poles, and compute their residues, along with assuring the validity of certain criteria and whatnot. And, to be very honest, although incredibly useful, I was never a fan of the laborious contour integration. So, it came to me as God’s gift when I learned DUTIS can often be used as a substitute instead.

If you’re like me, this should be melody to your ears. But I’m sure you won’t believe it until you see it. So, let’s get our hands dirty once more with the following integral:

While this integral is difficult because the integrand does not have an antiderivative in terms of elementary functions, it is possible to take the complex definition of the sine function and perform some elementary contour integration. However, why go through the trouble when Feynman’s trick comes to the rescue?

First we notice that the integral is symmetrical about the origin so it suffices to integrate over the positive real line and multiply by two. That is:

To begin solving this, however, we introduce not just a dummy variable but a dummy factor (this is legal because for a fixed value of the variable, we are left with the original :

We introduce a negative sign in the exponent to ensure convergence between the bounds of integration. Then, upon applying Feynman’s trick, we have:

Now the integral on the RHS can be evaluated easily by applying integration by parts twice in a row. We see then that our work is reduced to solving this:

Upon solving this differential equation, we are left with:

Where C is the constant of integration. Now, to solve for the constant of integration, we consider the integral as a → ∞. It is immediately apparent that the integrand and the integral tend to 0. And we see that C = π. Then, our work is trivialized again. We see that

Evaluating J(0), we see that J = π. And the integral is solved. No need for contour integration. No need for any sort of complications. DUTIS has saved the day once again!

With this, this article reaches its conclusion. I direct the interested reader to learn more about DUTIS with variable limits and in higher dimensions here.

Thanks for reading and I hope you have a great day!