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.

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.
