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?

Daniel Fischer
  • 199,211
  • 18
  • 249
  • 374
  • 24,247
  • 10
  • 91
  • 130
  • 48
    Strange, shouldn't it say "imaginary part" instead of "complex part"? – joriki Dec 09 '12 at 19:32
  • 17
    Although finding that specific integral may be hopeless, perhaps we can make one ourselves? Feynman was indeed a genius, but seeing how we would solve integrals involving branch cuts and such without use of any contour integration is interesting. You need not learn higher level of integration or of mathematics to do it; such is the meaning and application of counter integration. – AXH Dec 12 '12 at 00:53
  • 5
    Is this different from question 167304? – daniel Jan 18 '13 at 23:02
  • 11
    @downvoter: Why? – Argon Jan 19 '13 at 21:57
  • 39
    Check this out: http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf – Berkheimer Jan 23 '13 at 17:33
  • 1
    Thanks for posting the PDF, although it's kept me up tonight as I couldn't stop reading :) – Morten Jensen Mar 21 '13 at 07:18
  • 2
    Also see http://lesswrong.com/lw/cvf/where_fermi_fails_what_is_hard_to_estimate/ – Joel May 12 '13 at 00:57
  • 1
    @Berkheimer, would you please update the link? – MathArt Aug 28 '20 at 06:23
  • 1
    Updated link: http://fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf – Berkheimer Aug 28 '20 at 09:52

4 Answers4


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*}

  • 18,473
  • 2
  • 39
  • 92

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.

  • 634
  • 6
  • 13
  • 35
    because the tangent function is periodic of period $\pi$, to know what the value is you need to know $10^{100}\bmod\pi$; this is effectively the same as knowing $\pi$ to a hundred places. – Steven Stadnicki May 12 '13 at 01:00
  • 24
    It seems like Gleick didn't quite get the point. It's not enough to *know* the 100th digit of $\pi$. You have to divide $10^{100}$ by $\pi$ and calculate the remainder with a precision of something more than 100 decimal places. So even if you could produce the 100th digit of $\pi$ on short notice—or even if you could produce the first hundred-odd digits, which in fact are required—you still have to do the full 100-digit division in sixty seconds, which is impossible. Feynman explains this clearly in his memoir, but Gleick doesn't seem to understand. (I'm tempted to add "as usual".) – MJD Jun 20 '13 at 22:45
  • 1
    I think "The length of the Steiner tree on lattice points within the origin-centered sphere of volume 100" might be hard to solve in 1 minute. – Vladimir Reshetnikov Mar 22 '14 at 04:10
  • 8
    It's silly and pointless I know, but for reference, Mathematica tells me $10^{100} \mod \pi=0.381568$. –  Mar 31 '14 at 04:56
  • 2
    This story is also in *Surely You're Joking* (page 195 in my copy); it's right before the story about the contour integral. – Nate Eldredge May 26 '14 at 15:49
  • To provide a plausible $60$-second method for computing the binomial coefficient: Use Stirling's formula on the expression $20!/(10!^2)$. This gives $\frac{\sqrt{40\pi}}{\sqrt{20\pi}^2}\cdot \frac{(20/e)^{20}}{(10/e)^{20}}$, so we have $\frac1{\sqrt{10\pi}}\cdot 2^{20}$, which is roughly $\frac15\cdot 10^6$, so you answer $200000$, which is just $8\%$ over the correct answer of $184756$. (Recognizing that $\sqrt{10\pi}$ is more like $5.5$ would tell you to go down by around $10\%$, and give an even closer $180000$.) – RavenclawPrefect Jun 02 '21 at 20:55

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.

  • 1,617
  • 1
  • 14
  • 28

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.

Manjil P. Saikia
  • 542
  • 2
  • 13
  • 4
    The point is that, anyway, a good reference, detailed explanation, even if no one can give an exact reference to the integral, would be of great help. There is a comment with an mit link that has nice informations, maybe some contribution from each would help us figure out at least a good explanation. The point of the question is not just that Feynman was stumped, but that this, by all means, leads to very interesting calculus for all of us who have subscribed and upvoted the question. P.S. it's not my downvote, however I can agree that is not an answer worth the bounty. Best wishes, Mike – Mihai Bujanca Mar 19 '13 at 19:22
  • 1
    As the answer by Joel shows, this answer refers to a different situation than the question. – GeoffDS May 12 '13 at 00:49