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?

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    $\begingroup$ Strange, shouldn't it say "imaginary part" instead of "complex part"? $\endgroup$
    – joriki
    Commented Dec 9, 2012 at 19:32
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    $\begingroup$ 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. $\endgroup$
    – AXH
    Commented Dec 12, 2012 at 0:53
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    $\begingroup$ Is this different from question 167304? $\endgroup$
    – daniel
    Commented Jan 18, 2013 at 23:02
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    $\begingroup$ @downvoter: Why? $\endgroup$
    – Argon
    Commented Jan 19, 2013 at 21:57
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    $\begingroup$ Check this out: ocw.mit.edu/courses/mathematics/… $\endgroup$
    – Berkheimer
    Commented Jan 23, 2013 at 17:33

4 Answers 4


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*}

  • $\begingroup$ surely one of your Re's should be an Im? $\endgroup$
    – akkkk
    Commented Jun 24, 2014 at 22:50
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    $\begingroup$ @akkkk: Notice that $dz = i e^{i\phi} d\phi$. $\endgroup$
    – user26872
    Commented Jun 24, 2014 at 23:57

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.

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    $\begingroup$ 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. $\endgroup$ Commented May 12, 2013 at 1:00
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    $\begingroup$ 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".) $\endgroup$
    – MJD
    Commented Jun 20, 2013 at 22:45
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    $\begingroup$ 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. $\endgroup$ Commented Mar 22, 2014 at 4:10
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    $\begingroup$ It's silly and pointless I know, but for reference, Mathematica tells me $10^{100} \mod \pi=0.381568$. $\endgroup$
    – user18862
    Commented Mar 31, 2014 at 4:56
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    $\begingroup$ This story is also in Surely You're Joking (page 195 in my copy); it's right before the story about the contour integral. $\endgroup$ Commented May 26, 2014 at 15:49

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.

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    $\begingroup$ what is the reason for votedown? $\endgroup$
    – Manoj
    Commented Mar 26, 2013 at 12:22
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    $\begingroup$ This doesn't really answer the question. I am trying to find if anyone knows what the exact integral was. (I didn't downvote, however) $\endgroup$
    – Argon
    Commented Mar 27, 2013 at 18:24
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    $\begingroup$ I hope I had teacher like Feynman's. Quite a nice story, though :) $\endgroup$
    – Sawarnik
    Commented Feb 4, 2014 at 15:58

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.

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    $\begingroup$ 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 $\endgroup$ Commented Mar 19, 2013 at 19:22
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    $\begingroup$ As the answer by Joel shows, this answer refers to a different situation than the question. $\endgroup$
    – GeoffDS
    Commented May 12, 2013 at 0:49

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