It's a simple enough result I would have been unsurprised if it weren't named for anyone at all. I certainly find it odd it's named for a relatively modern physicist rather than an early-calculus mathematician. I assume Feynman earned the moniker by popularising its use in physics, but even that is a shock since it has obvious applications to 19th century thermodynamics.

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    $\begingroup$ For what it's worth we don't call it that in mathematics. $\endgroup$ Jun 20, 2017 at 19:06
  • $\begingroup$ Could you link or cite where it is so named? $\endgroup$
    – Conifold
    Jun 21, 2017 at 3:15

1 Answer 1


The name comes from Feynman's evaluation of the integral $$\int_0^\infty\frac{\sin x}{x}dx.$$ This is a classical example on applications of residues (I think it is originally due to Hardy. In the US residues are not a standard part of the undergraduate curriculum).

Feynman proposed the following elementary argument. Consider instead $$f(y):=\int_0^\infty\frac{\sin x}{x}e^{-xy}dx,$$ so we need $f(0)$. Then differentiate under the integral sign: $$f'(y)=-\int_0^\infty e^{-xy}\sin x dx,$$ and this is easily evaluated (integration by parts twice). As we know $f(y)\to 0$ as $y\to+\infty$ we find $f$ and our integral is $f(0)$.

This simple argument impressed people so much that they started calling it Feynman's trick. Those for whom this was the first interesting example of differentiation under the integral sign extended this name to any differentiation under integral sign.

  • $\begingroup$ So that term only applies to this very specific problem? $\endgroup$
    – Geremia
    Jun 20, 2017 at 19:55
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    $\begingroup$ I've never heard professional mathematicians using this term. Other people decide themselves what is the scope of "Feynman's trick". Feynman's books for general public are read by much larger set of people then professional mathematicians, and I think he mentions this trick in one of them. $\endgroup$ Jun 20, 2017 at 19:58
  • $\begingroup$ Excellent historical detail. FWIW $f=\arctan\frac{1}{y}$. $\endgroup$
    – J.G.
    Jun 20, 2017 at 21:15
  • $\begingroup$ Why do you consider it is due to this specific evaluation, and what is the evidence Feynman used it in this specific example? I have never read anything suggesting that before. What seems more convincing than any specific evaluation is Feynman's mention of the method (without explicit examples) in his book SYJMF. Only in the 21st century did I read that anyone used the name "Feynman's trick" (and only online, as you said not by professional mathematicians). $\endgroup$
    – KCd
    Jun 22, 2017 at 12:11
  • $\begingroup$ Update: A reference where Feynman computes the integral of $(\sin x)/x$ over $[0,\infty)$ is in the Hughes Lectures Vol. 5, pp. 13-14, at thehugheslectures.info/the-lectures. $\endgroup$
    – KCd
    Jan 4, 2019 at 1:26

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