# On upper bounds of the complex modulus of the Mellin transform of the cosine function, when is restricted to a given rectangle

The Mellin transform $$\left\{ M f\right\}(s)$$ of the cosine function $$f(x)=\cos(x)$$ was calculated as $$\int_0^\infty \cos(x)x^{s-1}dx=\cos\left(\frac{\pi s}{2}\right)\Gamma(s)$$ for $$0<\Re s<1$$, where $$\Gamma(s)$$ is the Gamma function.

I would like to learn more about this. I know the Maximum Modulus Principle of complex analysis, and I've asked what's about of an interesting upper bound of the modulus $$|\left\{ M f\right\}(s)|$$, when $$s$$ lies in a rectangle, for example $$\frac{1}{4}\leq \Re s\leq \frac{3}{4}$$ and with $$-L\leq \Im s\leq L$$, where $$L>0$$ is a real number. I know Stirling formula, for example see [1]. And from 11.14.33 of [2] combined with the definiton of the cosine function one has $$\lim_{\gamma\to\pm\infty}\frac{e^{(\sigma+i\gamma)i\pi/2}+e^{-(\sigma+i\gamma)i\pi/2}}{2}\Gamma\left(\sigma+i\gamma\right)=0,$$ where we are denoting with $$s=\sigma+i\gamma$$ the complex variable, and in our example this limit makes sense for $$0<\sigma<1$$.

Please only is required a detailed explanation of one of the following questions, well b) (if in next case b) is feasible to deduce something about the asymptotic behaviour of the complex modulus) or well a) and hints to get the other.

Question. What's about of an interesting upper bound of the modulus of our Mellin transform $$\left| \int_0^\infty \cos(x)x^{s-1}dx \right|$$

a) in a rectangle with sides defined by $$\frac{1}{4}\leq \Re s\leq \frac{3}{4}$$ and with $$-L\leq \Im s\leq L$$, where $$L>0$$ is a small real number;

b) in a rectangle with sides defined by $$\frac{1}{4}\leq \Re s\leq \frac{3}{4}$$ and with $$-L\leq \Im s\leq L$$, where $$L>0$$ is a very large real number, that is in this second exercise we are interested in the asymptotic behaviour of upper bouds for the complex modulus as $$L\to\infty$$. Many thanks.

## References:

[1] Garrett, Counting zeros of $$\zeta(s)$$. Lecture notes from University of Minnesota.

[2] Digital Library of Mathematical Functions. Section 1.14 Integral Transforms.

## 1 Answer

The easiest thing to do : $$\Gamma(\sigma+2i \pi \xi) = \int_0^\infty x^{\sigma+2i \pi \xi-1} e^{-x}dx \underset{x = e^{-u}}=\int_{-\infty}^\infty e^{-e^{-u}} e^{-\sigma u} e^{-2i \pi \xi u}du$$ is the Fourier transform of a Schwartz function $f_\sigma(u) = e^{-e^{-u}} e^{-\sigma u}$ whenever $\sigma > 0$.

When integrating by parts a Fourier transform $$(2i \pi \xi)^n\Gamma(\sigma+2i \pi \xi) = \int_{-\infty}^\infty (\frac{d^n}{d u^n}f_\sigma(u))e^{-2i \pi \xi u}du$$ Since $\frac{d^n}{d u^n}f_\sigma(u)$ is also a Schwartz function, and hence $L^1$, it means that $(2i \pi \xi)^n\Gamma(\sigma+2i \pi \xi)$ is bounded,

so that $\Gamma(\sigma+2i \pi \xi)$ decreases faster than any polynomial as $\xi \to \pm \infty$.

(more generally, the Schwartz functions are closed under Fourier transform)

Finally $\Gamma(s)= \frac{\Gamma(s+1)}{s}$ tells us this property stays true for $Re(s) \le 0$.

• Many thanks for your attention, now I am going to study the answer. – user243301 Jan 6 '17 at 10:50