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Let $c \in (1/2,3/2)$. In my research I have encountered the inverse Mellin transform of $\zeta(s-1)/s$:

$$f(x) = \frac{1}{2\pi i} \int_{(c)} \frac{\zeta(s-1)}{s}x^{-s} ds.$$ Does it have a closed form? If not, what can be said about it qualitatively, e.g. does it have constant sign? What are its asymptotics?

If $\zeta$ is replaced by $1$, this is the usual computation with Heaviside step function which is essential to Perron's formula.

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    $\begingroup$ if $\zeta(s-1)$ is replaced by $\zeta(s)$ this equals the integer part of $x$ --- math.stackexchange.com/q/751948/87355 $\endgroup$
    – Carlo Beenakker
    Commented May 8, 2021 at 12:18
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    $\begingroup$ Did you investigate the convergence? Once you know that for $c >2$, $\frac{1}{2\pi i}\lim_{T\to \infty} \int_{c-iT}^{c+iT} \frac{\zeta(s-1)}{s}x^s ds=\sum_{n\le x} n$ then apply the residue theorem to $\lim_{a\to 0^+} \frac{1}{2\pi i} \int_{(c)} e^{a(s-c)^2} \frac{\zeta(s-1)}{s}x^s ds$ to obtain $f(1/x)=\sum_{n\le x} n-\frac{x^2}2$ at least in the sense of distributions. $\endgroup$
    – reuns
    Commented May 8, 2021 at 13:01

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