What are some of the more unknown and/or underappreciated things that Euler discovered? The man has done so much that there's bound to be notable results that most people aren't aware of.

This could be, say, a constant or theorem that not many people are aware of, or possibly a well-known theorem that is named after someone else even though he was actually the first one to discover it.

I'd request that, in your answer, you explain what makes the result particularly remarkable despite its unfamiliarity to the general math public. This could be the wide range of applications, or the fact that Euler lived over $200$ years ago, etc.

Of course, please stay away from well-known results such as $e^{i \pi} = -1$ or $V - E + F = 2$

  • $\begingroup$ Somebody said that all (most of?) calculus theorems are misnamed, that they were first written down by Euler. $\endgroup$
    – vonbrand
    Mar 5, 2014 at 14:44
  • $\begingroup$ @vonbrand Perhaps Stigler's Law of eponymy? $\endgroup$
    – MCT
    Mar 22, 2014 at 1:58

1 Answer 1


It is well-known that Euler calculated the values of $\zeta(2n)$ for $n>0$. However, it is less known that he essentially discovered the functional equation of the Riemann zeta function. Euler was interested in the divergent sums

$$1+2^n+3^n + \dots$$

when $n>0$. He noticed that

$$\frac{t}{1+t} = t-t^2+t^3 - \dots$$

so, with $t=1$ we find $$1-1+1-1+\dots = \frac{1}{2},$$ an answer which is of course very silly if we take the left hand side literally. But $1/2$ is indeed the "average" of the partial sums on the left hand side, so maybe it is not completely silly. If we continue with this insanity we find:

$$t\frac{d}{dt}\frac{t}{1+t} = \frac{t}{(1+t)^2} = t - 2t^2 + 3t^3 - \dots$$

and therefore $$1-2+3-\dots = \frac{1}{4}.$$ In general, we have

$$\left(t\frac{d}{dt}\right)^n\frac{t}{1+t}\biggr|_{t=1} = 1-2^n+3^n-4^n+\dots.$$

On the other hand, if we put $t=e^x$, then $t \frac{d}{dt} = \frac{d}{dx}$, and $t=1$ becomes $x=0$:

$$\frac{d^n}{dx^n}\frac{e^x}{1+e^x}\biggr|_{x=0} = 1-2^n+3^n-4^n+\dots.$$

Now for $s>1$ a simple calculation shows that

$$1-2^{-s} + 3^{-s} - \dots = (1-2\times 2^{-s})\zeta(s),$$ hence in analogy, he defined $\zeta(-n)$ as

$$\zeta(-n):= (1-2^{n+1})^{-1} \frac{d^n}{dx^n}\frac{e^x}{1+e^x}\biggr|_{x=0}$$

For instance, for $n=1$ we find

$$\zeta(-1) = (1-4)^{-1} \frac{1}{4} = -1/12.$$

Anyways, it turns out that $e^x/(1+e^x)$ is essentially the generating function of the Bernoulli numbers, and after some simple manipulations, Euler obtained that for $n>1$,

$$\zeta(1-n) = -\frac{B_n}{n}.$$

(Notice in passing that this reveals the "trivial zeroes" of the Riemann zeta function (since the odd Bernoulli numbers vanish).) Hence, putting this together with his work on the values $\zeta(2n)$, Euler found the following explicit formulas:

$$\zeta(1-2n) = -\frac{B_{2n}}{2n}$$ $$\zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}.$$

This is already enough to conjecture the form the functional equation should take, and Euler had all of the ingredients for it. However, it seems that Riemann was the first to explicitly write it down (and to prove it).

(I picked this up from Hida's book Elementary theory of $L$-functions and Eisenstein series (which is, for the most part, far from being elementary...).)

  • $\begingroup$ @FelixMarin Thanks! $\endgroup$ Mar 6, 2014 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.