Let Mellin transform of a function $f(x):[1,\infty)\rightarrow\mathbb{C}$ defined by:
$$F(s)=\int_{1}^{\infty}f(x)x^{-s-1}dx$$
I am trying to prove the following:
If $F(s)$ has a simple pole at $s=\sigma+it$, then $f(x)\neq o(x^{\sigma})$.
My line of thought is, assume $f(x)=o(x^{\sigma})$. Since $F(s)$ has pole at $\sigma+it$, $F(\sigma+it)$ is asymptotic to $\frac{C}{(s-\sigma-it)}$ for some constant $C$. On the other hand let $|f(x)|<\frac{|C|}{2}x^{\sigma}$ for some $x>A$, the integral becomes:
$$\int_{1}^{A}f(x)x^{-s-1}dx+\int_{A}^{\infty}f(x)x^{-s-1}dx<O(1)+\frac{|C|}{2|(s-\sigma)|}$$
Compare the singular part, the coefficient does not match, contradiction.
Edit
The singular part of $|F(s)|$ is $\frac{|C|}{|s-\sigma-it|}$, while the singular part is also less than $\frac{|C|}{2|s-\sigma-it|}$, which is a contradiction.
Is the above reasoning correct?