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Let $F(x)$ satisfy

$$\begin{equation}\int\limits_0^\infty \vert F(x)\vert x^{c-1}dx<\infty\end{equation}\tag{1}$$

for some $c\in(a,b)$ with $a,b\in \mathbb{R}$ and $a<b$. Then, the Mellin transform $f(s)$ necessarily exists and represents an holomorphic function on strip $\Re(s)\in(a,b)$:

$$f(s)=\int\limits_0^\infty F(x) x^{s-1}dx\tag{2}$$

and the inversion is given by the inverse Mellin transform

$$F(x)=\frac{1}{2\pi i}\int\limits_{c-i\infty}^{c+i\infty}f(s)x^{-s}ds\tag{3}$$

$\bf{Question}$: Can integral $\int_{c-i\infty}^{c+i\infty}f(s)x^{-s}ds$ from eq. ($3$) converge only conditionally but not absolutely, given $F(x)x^{c-1}\in L^1 ( 0,\infty )$ as described by eq. ($1$)?

Any example or reference is welcome.

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