Mellin transform definition 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 $ ?
 A: 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 $0<\Re(s)<1$.
A: $\newcommand{\+}{^{\dagger}}%
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$\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}
