Enquiry on a claim in Titchmarsh. [closed]

Question

There is a claim on p.107 of Titchmarsh's ''The theory of the Riemann zeta function'', that if $0<a<b \leq 2a$ and $t>0$, then

the bound

$$\sum_{a <n <b} n^{-\frac{1}{2}-it} \ll (a/t)^{1/2} + (t/a)^{1/2} \tag{1}$$

follows from

$$\sum_{a <n <b} n^{-it} \ll t^{1/2} + at^{-1/2} \tag{2}$$

''by partial summation.'' Titchmarsh doesn't include any details of the proof, but from the partial (Abel) summation formula:

https://en.wikipedia.org/wiki/Abel%27s_summation_formula

it's not clear to me how (2) follows from (1). Can someone kindly fill in the details omitted by Titchmarsh?

Answer 1

Summation by parts has several variants. We use the following.

Define $a(n) = n^{-it}$ for $n \geq a$. Define its partial sum from $a$ by $A(u) = \sum_{a \leq n \leq u} a(n)$. Then Titchmarsh apparently showed that $$ A(u) = O(t^{1/2} + a t^{-1/2}) \qquad (u > a).$$ Partial summation (in the way coming directly from Riemann-Stieltjes integration) says that for a $C^1$ function $f$, we have $$ \sum_{a < n \leq b} a(n) f(n) = A(b)f(b) - A(a)f(a) - \int_a^b A(u) f'(u) du. $$ Apply this with $f(x) = x^{-1/2}$, which is $C^1$ for $x > 0$. Then we find $$ \sum_{a < n \leq b} n^{-\frac{1}{2} - it} = A(b)/b^{1/2} - A(a)/a^{1/2} + \frac{1}{2} \int_a^b A(u)/u^{3/2} du. $$ The integral is logarithmic in size, which Titchmarsh ignores. The initial sums are bounded in the big-Oh sense by $$ (t^{1/2} + at^{-1/2}) / b^{1/2} + n^{-it}/a^{1/2} \ll (t/b)^{1/2} + a/(tb)^{1/2} \ll (t/a)^{1/2} + (a/t)^{1/2}.$$ We've used that $b > a$ in the final simplification, though it's not obvious why such a symmetric form is desirable. I didn't look at the linked document to see what's done with it.