What are other methods to Evaluate $\int_0^{\infty} \frac{y^{m-1}}{1+y} dy$? I am looking for an alternative method to what I have used below. The method that I know makes a substitution to the Beta function to make it equivalent to the Integral I am evaluating.


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*Usually we start off with the integral itself that we are evaluating (IMO, these are better methods)  and I would love to know such a method for this. 

*Also, I would be glad to know methods which uses other techniques that I am not aware, which does not necessarily follow (1)

$$\Large{\color{#66f}{B(m,n)=\int_0^1 x^{m-1} (1-x)^{n-1} dx}}$$
$$\bbox[8pt,border: 2pt solid crimson]{x=\frac{y}{1+y}\implies dx=\frac{dy}{(1+y)^2}}$$
$$\int_0^{\infty} \left(\frac{y}{1+y}\right)^{m-1} \left(\frac{1}{1+y}\right)^{n-1} \frac{dy}{(1+y)^2}=\int_0^{\infty} y^{m-1} (1-y)^{-m-n} dy$$
$$\Large{n=1-m}$$
$$\Large{\color{crimson}{\int_0^{\infty} \frac{y^{m-1}}{1+y} dy=B(m,1-m)=\Gamma(m)\Gamma(m-1)}}$$
Thanks in advance for helping me expand my current knowledge.

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.
