# Why is differentiation under the integral sign named Feynman's trick?

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.

• For what it's worth we don't call it that in mathematics. – mathematician Jun 20 '17 at 19:06
• Could you link or cite where it is so named? – Conifold Jun 21 '17 at 3:15

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 $$f(0)$$.
• Excellent historical detail. FWIW $f=\arctan\frac{1}{y}$. – J.G. Jun 20 '17 at 21:15
• 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. – KCd Jan 4 '19 at 1:26