is thre a possible meaning or definition to the Mellin transform
$$ \int_{0}^{\infty} \frac{dt}{t-1}t^{s-1}= F(s) $$
i know that $$ \int_{0}^{\infty} \frac{dt}{t+1}t^{s-1}= F(s)= \frac{\pi}{sin(\pi s)} $$
howver can one overcome the pole at $ t=1 $ ?
Well, as it was mentioned, you can use the Cauchy principal value:
$$
\begin{eqnarray}
F(s)=\int_{0}^{\infty} \frac{dt}{t-1}t^{s-1}&=&
\lim_{\varepsilon\rightarrow 0+} \left[
\int_0^{1-\epsilon } \frac{t^{s-1}}{t-1} \, \mathrm dt+\int_{1+\epsilon}^{\infty } \frac{t^{s-1}}{t-1} \, \mathrm dt
\right]=
\\&=&\lim_{\varepsilon\rightarrow 0+} \left[-B_{1-\epsilon}(s,0)+B_{\frac{1}{\epsilon+1}}(1-s,0)\right]=\\
&=&\psi ^{(0)}(s)-\psi ^{(0)}(1-s)=\\
&=&-\pi \cot(\pi s)
\end{eqnarray}
$$
where $B_{x}(a,b)$ is the incomplete beta function and $\psi ^{(0)}(x)$ is the digamma function. The result is valid while $0<\Re(s)<1$.
Or you can modify your integral by splitting it into two parts:
$$
F(s)=\int_{0}^{\infty} \frac{dt}{t-1}t^{s-1}=
\int_0^1\frac{t^{s-1}}{t-1} \, \mathrm dt+\int_1^{\infty } \frac{t^{s-1}}{t-1} \, \mathrm dt
$$
After changing the variable in the second integral $x=\frac{1}{t}$ one gets
$$
F(s)=-\int_0^1\frac{t^{s-1}}{1-t} \, \mathrm dt-\int_0^{1 } \frac{x^{-s}}{1-x} \, \mathrm dx
$$
Combining them in one:
$$F(s)=-\int_0^1\frac{t^{s-1}+t^{-s}}{1-t} \, \mathrm dt=-\pi \cot(\pi s)$$
The last equality is taken from Gradshteyn and Ryzhik's "Table of Integrals, Series, and Products" and is valid while $0<\Re(s)<1$.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\displaystyle{? \equiv {\cal P}\int_{0}^{\infty} \frac{dt}{t-1}t^{s-1}\,, \quad 0 < s < 1}$.
We perform the integration $$ \int_{C}{z^{s - 1} \over z - 1}\,{\dd z \over 2\pi\ic}\quad\mbox{where}\quad z^{s - 1} = \verts{z}^{s - 1}\expo{\ic\pars{s - 1}\phi\pars{z}}\,,\quad z \not = 0,\quad 0 < \phi\pars{z} < 2\pi\tag{1} $$ There are not poles inside the integration contour $C$ such that $\ul{\mbox{the integral in $\pars{1}$ vanishes out}}$ !!!.
$C_{R}$ lies in a circle of radius $R$ while $C_{\epsilon}$ lies in a circle of radius $\epsilon$. In the limits $R \to \infty$ and $\epsilon \to 0^{+}$, the contribution of $C_{R}$ $\ds{\pars{~\sim {1 \over R^{1 - s}}~}}$ and $C_{\epsilon}$ $\ds{\pars{~\sim \epsilon^{s}~}}$ vanishes out since $\ds{0 < s < 1}$. \begin{align} 0&= \int_{C}{z^{s - 1} \over z - 1}\,{\dd z \over 2\pi\ic} = \overbrace{% \int_{0}^{\infty}{t^{s - 1} \over \left(t + {\rm i}0^{+}\right) -1}\,{\rm d}t} ^{\ds{\mbox{over}\ C_{+}}}\ +\ \overbrace{% \int_{\infty}^{0}{t^{s - 1} {\rm e}^{2\pi\left(s - 1\right){\rm i}} \over \left(t - {\rm i}0^{+}\right) - 1}\,{\rm d}t}^{\ds{\mbox{over}\ C_{-}}} \\[3mm]&= \int_{0}^{\infty}t^{s - 1} \left\lbrack{\cal P}{1 \over t - 1} - {\rm i}\,\pi\,\delta\left(t - 1\right)\right\rbrack \,{\rm d}t - \int_{0}^{\infty}t^{s - 1}{\rm e}^{2\pi s{\rm i}} \left\lbrack{\cal P}{1 \over t - 1} + {\rm i}\,\pi\,\delta\left(t - 1\right)\right\rbrack \,{\rm d}t \\[3mm]&= \left(1 - {\rm e}^{2\pi s{\rm i}}\right) {\cal P}\int_{0}^{\infty}{t^{s - 1} \over t - 1}\,{\rm d}t - {\rm i}\,\pi\left(1 + {\rm e}^{2\pi s{\rm i}}\right) \\[3mm]&\Longrightarrow \\[3mm]& {\cal P}\int_{0}^{\infty}{t^{s - 1} \over t - 1}\,{\rm d}t = {\rm i}\,\pi\, {1 + {\rm e}^{2\,\pi\,s{\rm i}} \over 1 - {\rm e}^{2\,\pi\, s\,{\rm i}}} = {\rm i}\,\pi\, {{\rm e}^{-\pi\,s\,{\rm i}} + {\rm e}^{\pi\,s\,{\rm i}} \over {\rm e}^{-\pi\,s{\rm i}} - {\rm e}^{\pi\,s\,{\rm i}}} = {\rm i}\,\pi\ {\phantom{-}2\cos\left(\pi s\right) \over -2{\rm i}\sin\left(\pi s\right)} = {\large -\pi\,{\rm cot}\left(\pi s\right)} \end{align}
