What are other methods to Evaluate $\int_0^{\infty} \frac{y^{m-1}}{1+y} dy$?
The integrand function behaves like $y^{m-2}$ on $[1,+\infty)$ and like $y^{m-1}$ in a right neighbourhood of zero, hence we must have $m-2<-1$ and $m>0$ in order to ensure integrability, so $m\in(0,1)$. In such a case we have: $$ I = \int_{0}^{+\infty}\frac{y^{m-1}}{1+y}\,dy = 2\int_{0}^{+\infty}\frac{y^{2m-1}}{1+y^2}\,dy $$ and since: $$ \int_{0}^{+\infty}\sin(u) e^{-yu}\,du = \frac{1}{1+y^2},$$ we have: $$ I = 2\int_{0}^{+\infty}\int_{0}^{+\infty} y^{2m-1} e^{-yu}\sin(u) \,dt\,du = 2\,\Gamma(2m)\int_{0}^{+\infty}\frac{\sin u}{u^{2m}}\,du $$ so the problem boils down to evaluating: $$ J(\alpha) = \int_{0}^{+\infty}\frac{\sin u}{u}u^{\alpha}\,du $$ for $\alpha=1-2m\in(-1,1)$. The last integral can be computed with standard complex analytic techniques (i.e. Mellin transform) and leads to: $$ J(\alpha) = \Gamma(\alpha)\sin\left(\frac{\pi\alpha}{2}\right), $$ so: $$ I = 2\Gamma(2m)\Gamma(1-2m)\cos(\pi m)=\frac{2\pi\cos(\pi m)}{\sin(2\pi m)}=\frac{\pi}{\sin(\pi m)}. $$
I think something might be wrong with your answer, suppose $m\geq 0$, let $v = 1+y$. Your integral becomes $$I=\int\limits_1^{\infty} \frac{(v-1)^m-1}{v}dv . $$ We can then apply the binomial expansion and transform the integral into $$I= \int\limits_1^{\infty} \frac{ \sum\limits_{i=0}^{m}\left[ {m \choose i} (-1)^iv^{m-i} \right]-1}{v}dv = \int\limits_1^{\infty} \sum\limits_{i=0}^{m}\left[ {m \choose i} (-1)^iv^{m-i-1} \right]-\frac1v dv . $$ This diverges for $m\geq 0$. However, your answer of $\Gamma(m)\Gamma(m-1)$ would lead us to believe that for negative $m$ the integral is undefined but $m\geq 1$ would be no problem. For instance if $m=4$ then $\Gamma(4)\Gamma(3)$ is defined, whereas your integral is not.
EDIT: With the updated question we may use the same method to obtain $$I = \int\limits_1^{\infty} \sum\limits_{i=0}^{m}\left[ {m \choose i} (-1)^iv^{m-i-2} \right] dv,$$ for integer $m \geq 1$, which does agree with the corrected result that $B(m,1-m) = \frac{\pi}{\sin(\pi m)}$, namely, it is undefined.
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With $\ds{\Re\pars{m} \in \pars{0,1}}$:
$\large\left. 1\right)$ \begin{align} \int_{0}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y & = \int_{0}^{\infty}y^{m - 1}\ \overbrace{\int_{0}^{\infty}\expo{-\pars{1 + y}x}\,\dd x}^{\ds{1 \over 1 + y}}\ \,\dd y = \int_{0}^{\infty}\expo{-x}\ \overbrace{\int_{0}^{\infty}y^{m - 1}\expo{-xy}\dd y} ^{\ds{x^{-m}\,\Gamma\pars{m}}}\ \,\dd x \\[5mm] & = \Gamma\pars{m}\int_{0}^{\infty}x^{-m}\expo{-x}\,\dd x = \bbx{\ds{\Gamma\pars{m}\Gamma\pars{-m + 1} = {\pi \over \sin\pars{\pi m}}}} \end{align}
$\large\left. 2\right)$ I'll consider the $\ds{z^{m - 1}}$ branch-cut along $\ds{\left[0,\infty\right)}$ with $\ds{0 < \,\mrm{arg}\pars{z} < 2\pi}$. The integral has a single pole at $\ds{\expo{\pi\ic}}$. Contributions from a small arc around the origen and a big arc vanish when 'usual' limits are taken:
\begin{align} \color{#88f}{\int_{0}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y} & = 2\pi\ic\pars{\expo{\pi\ic}}^{m - 1} - \int_{\infty}^{0}{y^{m - 1}\expo{2\pars{m - 1}\pi\ic} \over 1 + y}\,\dd y = -2\pi\ic\expo{m\pi\ic} + \expo{2m\pi\ic}\color{#88f}{\int_{0}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y} \\[5mm] \int_{0}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y & = -\,{2\pi\ic\expo{m\pi\ic} \over 1 - \expo{2m\pi\ic}} = {2\pi\ic \over \expo{m\pi\ic} - \expo{-m\pi\ic}} = {2\pi\ic \over 2\ic\sin\pars{\pi m}} = \bbx{\ds{{\pi \over \sin\pars{\pi m}}}} \end{align}
$\large\left. 3\right)$ ${\large The\ 'weird'\ one\ !!!}$ \begin{align} &\int_{0}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y = \int_{0}^{1}{y^{m - 1} \over 1 + y}\,\dd y + \int_{1}^{\infty}{y^{m - 1} \over 1 + y}\,\dd y = \int_{0}^{1}{y^{m - 1} \over 1 + y}\,\dd y - \int_{1}^{0}{y^{-m} \over 1 + y}\,\dd y \\[5mm] = &\ \int_{0}^{1}{y^{m - 1} + y^{-m} \over 1 + y}\,\dd y = \int_{0}^{1}{y^{m - 1} + y^{-m} - y^{m} - y^{-m + 1} \over 1 - y^{2}}\,\dd y \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{y^{m/2 - 1} + y^{-m/2 - 1/2} - y^{m/2 - 1/2} - y^{-m/2} \over 1 - y}\,\dd y \\[5mm] = &\ {1 \over 2}\bracks{% -\int_{0}^{1}{1 - y^{m/2 - 1} \over 1 - y}\,\dd y - \int_{0}^{1}{1 - y^{-m/2 - 1/2} \over 1 - y}\,\dd y + \int_{0}^{1}{1 - y^{m/2 - 1/2} \over 1 - y}\,\dd y + \int_{0}^{1}{1 - y^{-m/2} \over 1 - y}\,\dd y} \\[5mm] = &\ {1 \over 2}\bracks{% -\Psi\pars{m \over 2} - \Psi\pars{-\,{m \over 2} + {1 \over 2}} + \Psi\pars{{m \over 2} + {1 \over 2}} + \Psi\pars{-\,{m \over 2} + 1}} \\[5mm] & = {1 \over 2}\bracks{\Psi\pars{1 - {m \over 2}} - \Psi\pars{m \over 2}} - {1 \over 2}\braces{\Psi\pars{1 - \bracks{{m \over 2} + {1 \over 2}}} - \Psi\pars{{m \over 2} + {1 \over 2}}} \\[5mm] & = {1 \over 2}\,\pi\cot\pars{\pi\,{m \over 2}} - {1 \over 2}\,\pi\cot\pars{\pi\,\bracks{{m \over 2} + {1 \over 2}}} = {1 \over 2}\,\pi\bracks{\cot\pars{\pi\,{m \over 2}} + \tan\pars{\pi\,{m \over 2}}} \\[5mm] & = {1 \over 2}\,\pi\,{\cos^{2}\pars{\pi m/2} + \sin^{2}\pars{\pi m/2} \over \sin\pars{\pi m/2}\cos\pars{\pi m/2}} = \bbx{\ds{\pi \over \sin\pars{\pi m}}} \end{align}