Specific instance of Mellin Inversion In my number theory notes, I have the following integral formula
$$
\frac{1}{2\pi i}\int_{(2)}\frac{t^s}{s(s+1)...(s+r)}ds=\begin{cases}
\frac{1}{r!}(1-\frac{1}{t})^r & t\geq 1\\
0 & 0<t<1
\end{cases}
$$
where the notation $\int_{(2)}$ means integrate along the contour $2-it$ for $t\in \mathbb{R}$. However, I do not know how this is proven as we only did the case when $r=1$. Although, it looks like it is of the form of Mellin inversion. Thus, my thought is if we let
$$
f(t)=\begin{cases}
\frac{1}{r!}(1-\frac{1}{t})^r & t\geq 1\\
0 & 0<t<1\end{cases}
$$
Then the above formula by Mellin inversion should be equivalent to showing that $(Mf)(s)=\frac{t^{2s}}{s(s+1)...(s+r)}$. However, I am getting stuck as when I take the Mellin transform of $f$, I have done the following using the binomial theorem
\begin{align*}
(Mf)(s)&=\int_1^\infty \frac{1}{r!}(1-\frac{1}{t})^rt^{s-1}dt\\
&=\int_1^\infty \frac{t^{s-1}}{r!}\sum_{i=0}^r\binom{r}{i}(-\frac{1}{t})^idt\\
&=\sum_{i=0}^r \frac{1}{i!(r-i)!}\int_1^\infty (-1)^it^{s-i-1}dt\\
&=\sum_{i=0}^r\frac{1}{i!(r-i)!}\left[(-1)^i\frac{t^{s-i}}{s-i}\right]_1^\infty\\
\end{align*}
However, the reason I am confused is that assuming conditions on $s$ so that the integral converges, I won't have a $t$ in the expression of $(Mf)(s)$, but I want for $(Mf)(s)=\frac{t^{s2}}{s(s+1)...(s+r)}$. Any help on understanding this integral would be much appreciated.
 A: This isn't a complete answer, but I believe there are a couple of clarifications that need to be made and it's a bit long for a comment.

Assuming
$$f_r(t)=\left\{\begin{array}{cc}
 \frac{\left(1-\frac{1}{t}\right)^r}{r!} & t\geq 1 \\
 0 & 0<t<1 \\
\end{array}\right.=\theta(t-1)\ \frac{\left(1-\frac{1}{t}\right)^r}{r!}\tag{1}$$
where $\theta(t-1)$ is the Heaviside step function and
$$F_r(s)=\frac{1}{\prod\limits_{n=0}^r (n+s)}=\frac{1}{s (s+1)_r}=\frac{\Gamma(s)}{\Gamma(r+s+1)}\tag{2}$$
where $(s+1)_r$ is the Pochhammer symbol, I believe the correct relationships are
$$\mathcal{M}_t[f_r(t)](-s)=\int\limits_0^\infty f_r(t)\ t^{-s-1}\,dt=\int\limits_1^\infty f_r(t)\ t^{-s-1}\,dt=F_r(s)\tag{3}$$
since $f_r(t)=0$ for $0<t<1$ and
$$\mathcal{M}_s^{-1}[F_r(s)]\left(\frac{1}{t}\right)=\frac{1}{2 \pi i} \int\limits_{a-i\,\infty}^{a+i\,\infty} F_r(s)\ \left(\frac{1}{t}\right)^{-s}\,ds=\frac{1}{2 \pi i} \int\limits_{a-i\,\infty}^{a+i\,\infty} F_r(s)\ t^s\,ds=f_r(t)\tag{4}$$

Note in formula (3) above the Mellin transform is evaluated at $-s$ instead of $s$, and in formula (4) above the inverse Mellin transform is evaluated at $\frac{1}{t}$ instead of $t$.

With the variable substitution $t=e^x$ the last integral in formula (3) above can be evaluated as the following Laplace transform.
$$\int\limits_1^\infty\frac{\left(1-\frac{1}{t}\right)^r}{r!}\ t^{-s-1}\,dt=\int\limits_0^\infty \frac{\left(1-\frac{1}{e^x}\right)^r}{r!}\ e^{-s x}\,dx=\mathcal{L}_x\left[\frac{\left(1-\frac{1}{e^x}\right)^r}{r!}\right](s)=\frac{\Gamma(s)}{\Gamma(r+s+1)}\tag{5}$$
See this Wolfram Alpha evaluation of the Laplace transform above.
