The Integral that Stumped Feynman?

Question

In "Surely You're Joking, Mr. Feynman!," Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods:

One time I boasted, "I can do by other methods any integral anybody else needs contour integration to do."

So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow.

Does anyone happen to know what this integral was?

Answer 1

I doubt that we will ever know the exact integral that vexed Feynman. Here is something similar to what he describes.

Suppose $f(z)$ is an analytic function on the unit disk. Then, by Cauchy's integral formula, $$\oint_\gamma \frac{f(z)}{z}dz = 2\pi i f(0),$$ where $\gamma$ traces out the unit circle in a counterclockwise manner. Let $z=e^{i\phi}$. Then $\int_0^{2\pi}f(e^{i\phi}) d\phi = 2\pi f(0).$ Taking the real part of each side we find $$\begin{equation*} \int_0^{2\pi} \mathrm{Re}(f(e^{i\phi}))d\phi = 2\pi \mathrm{Re}(f(0)).\tag{1} \end{equation*}$$ (We could just as well take the imaginary part.) Clearly we can build some terrible integrals by choosing $f$ appropriately.

Example 1. Let $\displaystyle f(z) = \exp\frac{2+z}{3+z}$. This is a mild choice compared to what could be done ... In any case, $f$ is analytic on the disk. Applying (1), and after some manipulations of the integrand, we find $$\int_0^{2\pi} \exp\left(\frac{7+5 \cos\phi}{10+6\cos\phi}\right) \cos \left( \frac{\sin\phi}{10+6 \cos\phi} \right) d\phi = 2\pi e^{2/3}.$$

Example 2. Let $\displaystyle f(z) = \exp \exp \frac{2+z}{3+z}$. Then \begin{align*}\int_0^{2\pi} & \exp\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \cos\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) \\ & \times\cos\left( \exp\left( \frac{7+5 \cos \phi}{10+6 \cos \phi} \right) \sin\left( \frac{\sin \phi}{10+6 \cos \phi} \right) \right) = 2\pi e^{e^{2/3}}. \end{align*}

Answer 2

Coincidentally, I just happened to be reading "Genius: The Life and Science of Richard Feynman" a few weeks ago. On page 178, James Gleick writes:

At lunch one day, feeling even more ebullient than usual, he challenged the table to a competition. He bet that he could solve any problem within sixty seconds, to within ten percent accuracy, that could be stated in ten seconds. Ten percent was a broad margin, and choosing a suitable problem was hard. Under pressure, his friends found themselves unable to stump him. The most challenging problem anyone could produce was: Find the tenth binomial coefficient in the expansion of $(1 + x)^{20}$. Feynman solved that just before the clock ran out. Then Paul Olum spoke up. He had jousted with Feynman before, and this time he was ready. The demanded the tangent of ten to the hundredth. The competition was over. Feynman would essentially have had to divide one by $\pi$ and throw out the first one hundred digits of the results - which would mean knowing the one-hundredth decimal digit of $\pi$. Even Feynman could not produce that on short notice.

Answer 3

The Question was regarding Differentiation under the integral Rule.

The direct quotation from "Surely You're Joking, Mr. Feynman!" regarding the method of differentiation under the integral sign is as follows:

One thing I never did learn was contour integration. I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. One day he told me to stay after class. "Feynman," he said, "you talk too much and you make too much noise. I know why. You're bored. So I'm going to give you a book. You go up there in the back, in the corner, and study this book, and when you know everything that's in this book, you can talk again." So every physics class, I paid no attention to what was going on with Pascal's Law, or whatever they were doing. I was up in the back with this book: Advanced Calculus, by Woods. Bader knew I had studied Calculus for the Practical Man a little bit, so he gave me the real works—it was for a junior or senior course in college.

It had Fourier series, Bessel functions, determinants, elliptic functions—all kinds of wonderful stuff that I didn't know anything about. That book also showed how to differentiate parameters under the integral sign—it's a certain operation. It turns out that's not taught very much in the universities; they don't emphasize it.

But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all their tools on it before giving the problem to me.

Answer 4

It was something related to calculating a definite integral that required Feynman to calculate some digits of $\pi$ after the decimal point. An exact reference to this integral would be very difficult to find if it exists in literature.