Why didn’t Euler conjecture the prime number theorem?


It shouldn’t have been too hard, based on what he knew. Let us take a look.

Euler knew that the sum of reciprocal primes diverged, and he even knew the growth rate of \sum_{p\le x} \frac{1}{p}, where the sum is taken only over primes p. His method of discovering this was rather elegant, even if his lack of respect for divergent series may look horrifying to a modern mathematician. It was already known earlier that the harmonic series grows like \log x; that is,

\sum_{1\le n\le x} \frac{1}{n}\approx \log x,

or more precisely,

\lim_{x\to\infty} \left(\sum_{1\le n\le x} \frac{1}{n}-\log x\right)=\gamma\approx 0.577.

Euler had earlier worked out the famous Euler product for the zeta function:

\sum_{n=1}^\infty \frac{1}{n^s} = \prod_p \left(1-\frac{1}{p^s}\right)^{-1}

for \Re(s)>1. (This formula has a delightful interpretation as an analytic formulation of the fundamental theorem of arithmetic.) Since neither side of the above formula makes sense when s=1, Euler did the most logical thing possible and set s=1 anyway, to obtain

1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \left(1-\frac{1}{2}\right)^{-1}\left(1-\frac{1}{3}\right)^{-1}\left(1-\frac{1}{5}\right)^{-1}\left(1-\frac{1}{7}\right)^{-1}\cdots.

Taking logarithms, he then obtained

\log\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)=-\log\left(1-\frac{1}{2}\right)-\log\left(1-\frac{1}{3}\right)-\log\left(1-\frac{1}{5}\right)-\log\left(1-\frac{1}{7}\right)-\cdots,

and expanding out all the logarithmic terms on the right as power series and regrouping, the right side becomes

\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right) + \frac{1}{2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots\right) + \frac{1}{3}\left(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{5^3}+\frac{1}{7^3}+\cdots\right)+\cdots.

Since all the terms except the first add up to something finite, they will not make a significant contribution in the limit, so Euler ignored them. (He could also have truncated after the first term when expanding the logarithms as power series, which would have had the same effect.) Hence, in the limit,

\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots = \log\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right) = \log\log\infty.

Naturally, that should be interpreted as

\sum_{p\le x} \frac{1}{p}\approx \log\log x.

One can get rid of all the divergent series and weird expressions like \log\log\infty by taking limits and being careful with error terms like a modern mathematician would do. But that wasn’t a necessary part of the culture in the 18th century.

At this point, Euler stopped, but I don’t understand why. It is now possible to make an estimate for the number of primes up to x, using a quick and easy computation. Since I’m a rebellious mathematician, I did the following computation on the front of an envelope.

Let us try to estimate the number of primes between x and kx, for some k>1, and let us write \pi(x) for the number of primes up to x. Then

\sum_{x\le p\le kx} \frac{1}{p}\approx \log\log kx - \log\log x\approx \frac{\pi(kx)-\pi(x)}{x},

where the explanation for the last term is that, if k is not much bigger than 1, then each summand on the left is roughly \frac{1}{x}. The goal is now to determine \pi(kx)-\pi(x). We have

\pi(kx)-\pi(x)\approx x(\log\log kx-\log\log x)=x\log\left(1+\frac{\log k}{\log x}\right).

Expanding the logarithm using a power series and truncating after the first term, we obtain

\pi(kx)-\pi(x)\approx \frac{x\log k}{\log x}.

Now, choosing k=1+\frac{1}{x} and again taking the first term of the power series for the logarithm (that seems to be the name of the game today), we obtain

\pi(x+1)-\pi(x)\approx \frac{1}{\log x}.

Now, summing over x gives

\sum_{x=1}^{n-1} (\pi(x+1)-\pi(x))=\pi(n)\approx \sum_{x=2}^{n-1}\frac{1}{\log x}\approx\int_2^n \frac{dt}{\log t}

(where I wrote 2 instead of 1 as the lower bound for the sum and integral to make them converge). And that’s one version of the prime number theorem. Nothing here would have caused Euler to break a sweat. But yet Gauss seems to have been the first one to conjecture the prime number theorem, and even he apparently did so on the basis of actually working out tables of primes and counting.

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About Simon

Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.
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3 Responses to Why didn’t Euler conjecture the prime number theorem?

  1. Maks says:

    I think you probably meant “take k = (x+1)/x” rather than “take k = 1/x”? Also in the last summation it should be 1/\log x instead of \log x. Otherwise, very nice!

  2. Xiannan Li says:

    This is nice! To be fair to Euler, the strength of the derivation lies in the information on $\sum_{x \le p \le kx} \frac 1p$, for which you used an exact formula that implicitly assumes the primes are very regularly distributed. This impressive regularity is obvious in hindsight (and as you’ve shown is equivalent to PNT), but probably Euler had very little reason to believe it.

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