Titchmarsh wrote that $$\zeta(s)=s\int_{1}^{\infty}\frac{\left \lfloor x \right \rfloor-x+\frac{1}{2}}{x^{s+1}}\,\mathrm{d}x+\frac{1}{s-1}+\frac{1}{2}\tag{2.14}$$ using the Euler-Maclaurin summation, for all $\Re s > 1$. Since the integral converges absolutely for all $\Re s > 0$, this equation now holds for $\Re s > 0$ and $s\neq 1$.
But below says that this integral converges for all $\Re s > -1$. I do not understand why. If you rewrite this integral using the integration by parts the new integral converges absolutely for all $\Re s > -1$, but how does this imply that it goes same for the original integral? Or is there something I am missing?
Let $s=\sigma+it$. Let $f(x)=\left \lfloor x \right \rfloor-x+\frac{1}{2}$ and $f_1(x):=\int_{1}^{x}f(y)\,\mathrm{d}y$.
Actually (2.1.4) gives the analytic continuation for $\zeta(s)$ for $\sigma>-1;$
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Hence $$\int_{x_1}^{x_2} \frac{f(x)}{x^{s+1}}\,\mathrm{d}x=\left [ \frac{f_1(x)}{x^{s+1}} \right ]_{x_1}^{x_2}+(s+1)\int_{x_1}^{x_2} \frac{f_1(x)}{x^{s+2}}\,\mathrm{d}x,$$ which tends to $0$ as $x_1\to\infty$, $x_2\to\infty$, if $\sigma>-1$. Hence the integral in (2.1.4) is convergent for $\sigma>-1$.