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.

  • $\begingroup$ 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:) ) $\endgroup$ – C Marius Jul 24 '17 at 8:25
  • $\begingroup$ 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 ... $\endgroup$ – C Marius Jul 24 '17 at 8:31
  • $\begingroup$ @CMarius I'm not using Morera's theorem, I'm proving $F$ is complex differentiable. $\endgroup$ – reuns Jul 24 '17 at 19:46
  • $\begingroup$ Look at $\int_1^\infty (x-\lfloor x \rfloor)^2 x^{-s-1}dx$ $\endgroup$ – reuns Jul 24 '17 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.