TITLE: What are other methods to Evaluate $\int_0^{\infty} \frac{y^{m-1}}{1+y} dy$?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}