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.


*

*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.
 A: 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:
$$ I = 2\Gamma(2m)\Gamma(1-2m)\cos(\pi m)=\frac{2\pi\cos(\pi m)}{\sin(2\pi m)}=\frac{\pi}{\sin(\pi m)}. $$
A: 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.
A: Here is an easy method using residues.
Substitute $y = x^2$ to turn the integral into
$$
I=\frac 1 2\int_0^\infty \frac{x^{2m-2}}{1+x^2} dx.
$$
(This substitution is optional but makes it a bit easier, in my opinion)
Therefore we need to calculate
$$
J = \int_0^\infty \frac{x^{a-1}}{1+x^2} dx.
$$
Consider the integral of the function $f(z) = \dfrac{z^{a-1}}{1+z^2}$ around a contour consisting of the real axis and an infinite semicircle in the upper half plane. The branch cut for $\log z$ we choose along the negative imaginary axis. This means that the argument equals $0$ along the positive real axis and $\pi$ along the negative real axis.
We need to make an infinitesimal indentation of the contour around the branch point zero.
The integrals along the indentation and the infinite semicircle are easily shown to vanish.
The integral from $0$ to $\infty$ equals $J$.
The integral from $-\infty$ to $0$ equals 
$$
-\int_0^\infty \frac{\left(e^{i \pi}x\right)^{a-1}}{1+x^2} dx=e^{i \pi a} J
$$
There is a residue equal to $\dfrac{\left(e^{i \pi/2}\right)^{a-1}}{2i} = \frac 1 2 e^{i \pi a/2}$ for the pole at $z = i$.
Applying the residue theorem gives the desired result
$$
J = \frac{\pi}{2 \sin{(\pi a/2)}}
$$
so
$$
I = \frac{\pi}{\sin{(\pi m)}}.
$$
If we had not made the first substitution, we would have had to use a keyhole contour.
