Domain of holomorphy and upper bound for a certain function defined by integral

Question

I have few questions about a certain function. Let
$$ g(\alpha) = \int_{\frac{1}{2}}^{\infty} \frac{\arcsin(\cos(\pi x)^2)^2}{x^{\alpha+2}} dx$$ Then

1. Is $g(\alpha)$ holomorphic? If so, then what is it's domain of holomorphy?
2. Can some one provide a sharp upper bound for $|g(\alpha)|$?

Here are my thoughts so far:

1. Should I use Cauchy-Riemann conditions ?!
2. $\arcsin(\cos(\pi x)^2)^2 < \frac{\pi^2}{4}$ hence $$|g(\alpha)| \leq \frac{\pi^2}{4} \int_{\frac{1}{2}}^{\infty} \frac{1}{|x^{\alpha+2}|} = \frac{\pi^2}{4} \frac{2^{\Re{\alpha}+1}}{\Re{\alpha} + 1}$$

Answer 1

Of course it is holomorphic for $\Re(\alpha)$ large enough.

If $F(s) = \int_c^\infty f(t) e^{-st}dt$ converges absolutely for $\Re(s) > \sigma$ then $G(s) = \int_c^\infty f(t) (-t) e^{-st}dt$ converges absolutely for $\Re(s) > \sigma$.

Thus for $\Re(s_0),\Re(s) > \sigma$ $$\int_{s_0}^s G(z)dz = \int_{s_0}^s(\int_c^\infty f(t) (-t) e^{-zt} dt)dz= \int_c^\infty f(t) (\int_{s_0}^s -t e^{-zt} dz)dt = F(s)-F(s_0)$$ proving $F$ is complex differentiable (holomorphic) and hence analytic.

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Then show for $\Re(s) > 1$ $$\zeta(s) = s \int_1^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1}+s \int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$$ $$=\frac{s}{s-1}-\frac{1}{2}+s \int_1^\infty (\lfloor x \rfloor-x+\frac{1}{2}) x^{-s-1}dx$$

Integrating by parts, you'll find a relation with your function.