# On a certain representation of the Riemann zeta function

Let $$\zeta$$ denote the Riemann zeta function. In this answer: https://mathoverflow.net/a/314066/133634, @Paul Garret considers the representation

$$\frac{\zeta(s)}{s} = \int_1^\infty (\sum_{1 \le n \le x} 1) x^{-s-1}dx$$ for $$\Re(s)>1$$, and then claims (without proof) that this representation yields

$$\frac{\zeta(s)}{s}+\frac{\zeta(s)}{1-s}= \int_0^\infty ((\sum_{1 \le n \le x} 1-x)-x(\sum_{1 \le n \le x} \frac1n-\log x-\gamma)) x^{-s-1}dx$$ for $$\Re(s)\in(0,1)$$, where $$\gamma$$ is the Euler constant.

But i can't see how the second representation follows from the first. Can anyone please explain ?

• Actually, it was @reuns who composed that answer. The derivation is pretty standard, in any case. – paul garrett Dec 19 '18 at 22:45
• @Paul Garret, any reference of the ''pretty standard'' derivation ? – OneTwoOne Dec 19 '18 at 23:11
• And it seems to me that reuns'representation is for $\frac{\zeta(s+1)}{s}$, for an unspecified radius of convergence. – OneTwoOne Dec 19 '18 at 23:36
• Titchmarsh? Edwards' book? Patterson's? Iwaniec'? Iwaniec-Kowalski? Karatsuba-Voronin? Ivic? Even Whittaker-and-Watson? I'd bet a dollar that most of these, if not all, include elementary relations like the above. The only thing that might be not entirely elementary is that the Euler-Mascheroni constant does give the constant in the asymptotics of the harmonic sum...? – paul garrett Dec 20 '18 at 0:02

Let $$0<\Re(s)<1$$ unless otherwise indicated. Then $$\int_0^\infty (\sum_{1 \le n \le x} 1-x) x^{-s-1}\,dx=-\int_0^1 x^{-s}+\int_1^\infty (\sum_{1 \le n \le x} 1-x) x^{-s-1}\,dx.$$ On the right hand side, the first integral equals $$1/(s-1)$$. The second integral equals $$\zeta(s)/s-1/(s-1)$$ as can be seen by proving it first for $$\Re(s)>1$$ and then continuing it analytically to $$\Re(s)>0$$. Hence $$\int_0^\infty (\sum_{1 \le n \le x} 1-x) x^{-s-1}\,dx=\frac{\zeta(s)}{s},\qquad 0<\Re(s)<1.\tag{1}$$ Now consider $$S(x):=\sum_{1 \le n \le x} \frac1n-\log x-\gamma,\qquad x>0.$$ Note that $$S(x)\ll\log x$$ when $$x<1$$, and $$S(x)\ll 1/x$$ when $$x\geq 1$$. Then, the remaining part of the integral is, by integration by parts, $$\int_0^\infty S(x)x^{-s}\,dx=\frac{1}{s-1}\int_0^\infty x^{1-s}\,dS(x).$$ Now we argue for the integral on the right hand side similarly as before: $$\int_0^\infty x^{1-s}\,dS(x)=\int_0^{1-} x^{1-s}\,dS(x)+\int_{1-}^\infty x^{1-s}\,dS(x).$$ On the right hand side, the first integral equals $$1/(s-1)$$. The second integral equals $$\zeta(s)-1/(s-1)$$ as can be seen by proving it first for $$\Re(s)>1$$ and then continuing it analytically to $$\Re(s)>0$$. Hence $$\int_0^\infty x^{1-s}\,dS(x)=\zeta(s),\qquad 0<\Re(s)<1,$$ and therefore $$\int_0^\infty S(x)x^{-s}\,dx=\frac{\zeta(s)}{s-1},\qquad 0<\Re(s)<1.\tag{2}$$ Subtracting $$(2)$$ from $$(1)$$, the claimed identity follows.