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Whenever I am not doing anything, I generally happen to see pages of some good Mathematical Institutes in India, so as to know more about the faculty members and see what they are working on.

While doing so, in one of the faculty webpage he has written this statement.

  • The prime number theorem has an one-line proof : The Riemann zeta function does not vanish on the one line "real part of $s$ equals 1".

Can anyone explain more about this proof. Is he assuming the Riemann Hypothesis? I am curious to know.

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    Check out [Shakarchi and Stein](http://books.google.com/books?id=0ECHh9tjPUAC&printsec=frontcover&dq=shakarchi+and+stein&hl=en&ei=MzMrTY-RD4Kclgf3suy8AQ&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCkQ6AEwAQ#v=onepage&q&f=false) volume 2, chapter 7.2011-01-10

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The assertion you made about $\zeta(s)$ does not assume the Riemann hypothesis and can, in fact, be proved reasonably quickly using reasonably elementary manipulations of trigonometric functions. The actual deduction of the prime number theorem from this is not trivial, and can be done (edit: as Matt E observes below, this is not the only -- and, indeed, not the original -- way) by some sort of "Tauberian" theorem: in these notes by Ash (which are based off a short paper by Donald Newman), the argument is to express \zeta'(s)/\zeta(s) as the "Mellin transform" (basically, a variant of the Laplace transform) of partial sums of the von Mangoldt function, denoted $\psi$. The point of the "Tauberian" theorem is that you can deduce properties about the original function from the Mellin transform: since we have some information about the Mellin transform (namely, because $\zeta$ is zero-free on the line $Re(s)=1$), we can deduce some information about $\psi$. This is enough to imply the prime number theorem.

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    ... in deriving the explicit formula. Regards,2011-09-08