In this article we obtain upper bounds for $\zeta(s)$ in the strip $\delta \leq \sigma \leq 2$, where $\delta > 0$. We will first show that for a fixed $0 < \varepsilon < \delta \leq 1$ we have
\[
\zeta(s) \ll \tau^{1- \delta + \varepsilon}
\] for $\delta \leq \sigma \leq 2$ and $|t| \geq 1$, where $\tau = |t| + 2$ and the implicit constant depends on $\delta$ and $\varepsilon$. Later we will establish a slightly stronger bound.
Note that for $\sigma > 1$ we have
\[ \begin{align} \zeta(s) = \sum_{n = 1}^{\infty} \frac{1}{n^s} &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^s}- \int_{n}^{n + 1} \frac{du}{u^{s}} \right) + \int_{1}^{\infty} \frac{du}{u^s} \\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^s}- \int_{n}^{n + 1} \frac{du}{u^{s}} \right) + \frac{1}{s-1}. \end{align}
\] Now let
\[
\varphi_n(s) = \frac{1}{n^s}- \int_{n}^{n + 1} \frac{du}{u^s}.
\] Then we have
\[
\zeta(s)- \frac{1}{s- 1} = \sum_{n = 1}^{\infty} \varphi_n(s).
\] Observe that
\[
|\varphi_n(s)| = \left|\int_{n}^{n + 1} \left( \frac{1}{n^s}- \frac{1}{u^s} \right) \, du \right| = \left|\int_{n}^{n + 1} \int_{n}^{u} \frac{s}{v^{s + 1}} dvdu \right| \leq \frac{|s|}{n^{\sigma + 1}}
\] for $n \in \mathbb{N}$ and $\sigma > 0$. Due to this inequality the series $\sum_{n = 1}^{\infty} \varphi_n$ converges uniformly on compact subsets of the half plane $\sigma > 0$. Hence, $\sum_{n = 1}^{\infty} \varphi_n$ is holomorphic in the half plane $\sigma > 0$ as each $\varphi_n$ is an entire function.
We will now combine the inequality $|\varphi_n(s)| \leq |s|/ n^{\sigma + 1}$ together with the trivial inequality $|\varphi_n(s)| \leq 2/n^{\sigma}$. This is because the first inequality helps with convergence in the critical strip whereas the second inequality does not involve $s$ and hence helps us get a sub power of $|s|$ in the bound. Let $0 < \lambda < 1$. Then we have
\[
|\varphi_n(s)| = |\varphi_n(s)|^{\lambda} |\varphi_n(s)|^{1-\lambda} \leq \left( \frac{|s|}{n^{\sigma + 1}} \right)^{\lambda} \left( \frac{2}{n^{\sigma}} \right)^{1- \lambda} \leq \frac{2 |s|^{\lambda}}{n^{\sigma + \lambda}}.
\] If $\sigma + \lambda > 1$, then we have
\[
\left| \sum_{n = 1}^{\infty} \varphi_n(s) \right| \leq 2 |s|^{\lambda} \sum_{n = 1}^{\infty} \frac{1}{n^{\sigma + \lambda}}.
\] Let $0 < \varepsilon < \delta \leq 1$. We take $\lambda = 1 -\delta + \varepsilon$ so that $0 < \lambda < 1$ and for $\sigma \geq \delta$ the series to the right converges. Thus for $\sigma \geq \delta$ we have
\[
\left| \sum_{n = 1}^{\infty} \varphi_n(s) \right| \leq 2 |s|^{1 -\delta + \varepsilon} \sum_{n = 1}^{\infty} \frac{1}{n^{1 + \varepsilon}} \ll |s|^{1- \delta + \varepsilon},
\] where the implicit constant depends on $\delta$ and $\varepsilon$. Hence, it follows that
\[
\zeta(s) \ll \frac{1}{|s- 1|} + |s|^{1- \delta + \varepsilon} \ll |s|^{1- \delta + \varepsilon}
\] for $\sigma \geq \delta$, $|t| \geq 1$. For $\delta \leq \sigma \leq 2$ we have $|s| \ll \tau$ and so for fixed $0 < \varepsilon < \delta \leq 1$ we find that
\[
\zeta(s) \ll \tau^{1- \delta + \varepsilon}
\] for $ \delta \leq \sigma \leq 2, |t| \geq 1$, where the implicit constant depends on $\delta$ and $\varepsilon$.
Let $\delta > 0$ be fixed. We now show that
\[
\zeta(s) \ll (1 + \tau^{1- \sigma}) \min \left( \frac{1}{|\sigma- 1|}, \log \tau \right)
\] uniformly for $\delta \leq \sigma \leq 2$, $|t| \geq 1$. It is easily seen by Abel’s summation by parts formula that
\[
\zeta(s) = \sum_{n \leq x} n^{-s} + \frac{x^{1-s}}{s-1} + \{x\} x^{-s} -s \int_{x}^{\infty} \{u\} u^{-s-1} \, du
\] for $\sigma > 0$ and $s \neq 1$. We can bound the integral to the right as
\[
\int_{x}^{\infty} \{u\} u^{-s-1} \, du \ll \int_{x}^{\infty} u^{- \sigma- 1} \, du = \frac{x^{- \sigma}}{\sigma}
\] for $\sigma > 0$. Thus we find that
\[
\zeta(s) = \sum_{n \leq x} n^{-s} + \frac{x^{1-s}}{s-1} + O(\tau x^{-\sigma}) \tag{1}
\] uniformly for $\sigma \geq \delta > 0$, $s \neq 1$. The key idea behind obtaining our next bound is to split the ranges of $s$ depending on the proximity of $\sigma$ to $1$.
Note that we have
\[
\left| \sum_{n \leq x} n^{-s} \right| \leq \sum_{n \leq x} n^{- \sigma} \leq 1 + \int_{1}^{x} u^{-\sigma} \, du
\] for $\sigma \geq 0$ and $x \geq 2$. If $0 \leq \sigma \leq 1 -\log x$, then
\[
\sum_{n \leq x} n^{-s} \ll 1 + \frac{x^{1- \sigma}}{1- \sigma} \ll \frac{x^{1- \sigma}}{1- \sigma},
\] where the constant $1$ gets absorbed as $x^{1- \sigma}/(1- \sigma) \geq 1$ for $0 \leq \sigma \leq 1- 1/\log x$. If $|\sigma- 1| \leq 1/ \log x$, then $1- 1/\log x \leq \sigma \leq 1 + 1/\log x$ and so we have
\[
e^{-1} u^{-1} \leq u^{-1 -1/\log x} \leq u^{-\sigma} \leq u^{-1 + 1/\log x} \leq e u^{-1}
\] for $1 \leq u \leq x$. Thus $u^{-\sigma} \asymp u^{-1}$ uniformly for $1 \leq u \leq x$. Using this we find that
\[
\sum_{n \leq x} n^{-s} \ll 1 + \int_{1}^{x} u^{-1} \, du \ll \log x.
\] If $1 + 1/\log x \leq \sigma \leq 2$, then
\[
\sum_{n \leq x} n^{-s} \ll 1 + \int_{1}^{\infty} u^{-\sigma} \, du = 1 + \frac{1}{\sigma- 1} \ll \frac{1}{\sigma- 1}.
\]
We now need to show that in each of the ranges of $\sigma$ the desired estimate holds. If $0 \leq \sigma \leq 1- 1/\log x$, then $|\sigma- 1| \geq 1/ \log x$ and so $1/|\sigma- 1| \leq \log x$. Thus $x^{1- \sigma}/(1- \sigma) \ll (1 + x^{1- \sigma}) \min (1/|\sigma- 1|, \log x)$ for $0 \leq \sigma \leq 1 -1/\log x$.
If $|\sigma- 1| \leq 1/\log x$, then $\log x \leq 1/|\sigma- 1|$ and so $\log x \ll (1 + x^{1- \sigma}) \min (1/|\sigma- 1|, \log x) $ for $|\sigma- 1| \leq 1/\log x$.
Finally, if $1 + 1/\log x \leq \sigma \leq 2$, then $|\sigma -1| \geq 1/\log x$ and $1/|\sigma- 1| \leq \log x$. Hence, $1/|\sigma- 1| \ll (1 + x^{1- \sigma}) \min (1/|\sigma- 1|, \log x)$ for $1 + 1/\log x \leq \sigma \leq 2$.
Now suppose that $|t| \geq 1$ and $\delta \leq \sigma \leq 2$. By taking $x = \tau$ in (1) we find that
\[
\zeta(s) \ll \sum_{n \leq x} n^{-s} + \tau^{1- \sigma} \ll (1 + \tau^{1- \sigma}) \min \left( \frac{1}{|\sigma- 1|}, \log \tau \right).
\] as $1/|s- 1| \leq 1$ due to $|t| \geq 1$ and $\min \left( 1/|\sigma- 1|, \log \tau \right) \gg 1$ for $0 \leq \sigma \leq 2$.