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

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 $$|g(\alpha)| \leq \frac{\pi^2}{4} \int_{\frac{1}{2}}^{\infty} \frac{1}{|x^{\alpha+2}|} = \frac{\pi^2}{4} \frac{2^{\Re{\alpha}+1}}{\Re{\alpha} + 1}$$

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.

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.

• Sir thank you very much for your answer! Let me see if I understand: for proving the holomorphy you use Morera's theorem but on (it seems to me, for c = 0) sort of Laplace transform of $f(t)$. Am I suppose to do the same thing with my function? Very good advice! (I forgot about that holomorphy criterion:) ) – C Marius Jul 24 '17 at 8:25
• Then, yes my function has some connection with $\zeta$ function, but how can it be proven form that known formula? I obtained the relation by a very different way ... – C Marius Jul 24 '17 at 8:31
• @CMarius I'm not using Morera's theorem, I'm proving $F$ is complex differentiable. – reuns Jul 24 '17 at 19:46
• Look at $\int_1^\infty (x-\lfloor x \rfloor)^2 x^{-s-1}dx$ – reuns Jul 24 '17 at 19:46