calculus - 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.

1. 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.


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

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$$\Large{\color{#66f}{B(m,n)=\int_0^1 x^{m-1} (1-x)^{n-1} dx}}$$

$$\bbox[8pt,border: 2pt solid crimson]{x=\frac{y}{1+y}\implies dx=\frac{dy}{(1+y)^2}}$$

$$\int_0^{\infty} \left(\frac{y}{1+y}\right)^{m-1} \left(\frac{1}{1+y}\right)^{n-1} \frac{dy}{(1+y)^2}=\int_0^{\infty} y^{m-1} (1-y)^{-m-n} dy$$

$$\Large{n=1-m}$$

$$\Large{\color{crimson}{\int_0^{\infty} \frac{y^{m-1}}{1+y} dy=B(m,1-m)=\Gamma(m)\Gamma(m-1)}}$$

Thanks in advance for helping me expand my current knowledge.

Answer

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:
$$I = 2\Gamma(2m)\Gamma(1-2m)\cos(\pi m)=\frac{2\pi\cos(\pi m)}{\sin(2\pi m)}=\frac{\pi}{\sin(\pi m)}.$$