Evaluation of integral related to Gamma function In few weeks ago, I encounter some integral. At that moment, it does not require specific value. But now I have a question on evaluation. Wolframalpha suggests the following answer.
$$
\int_{0}^{\infty}\frac{e^{-t}-1}{t^{a}}dt=\Gamma\left(1-a\right),\quad1<a<2.
$$
Well, by considering the singularity at $t=0$, it is easy to see that the integral has value.
But I don't have any experience on gamma function with negative evaluation. How can I verify that identity?
 A: For $t > 0$, let $f_t(s) = e^{-st}$. Then
$$e^{-t} - 1 = f_t(1) - f_t(0) = \int_0^1 f_t'(s)\,ds = -t\int_0^1 e^{-st}\,ds.$$
Substituting that into the given integral, we get
$$\int_0^{\infty} \frac{e^{-t}-1}{t^a}\,dt = -\int_0^{\infty} \int_0^1 \frac{e^{-st}}{t^{a-1}}\,ds\,dt.\tag{1}$$
Since the integrand on the right hand side of $(1)$ is positive and continuous, we can interchange the order of integration, so we get
\begin{align}
\int_0^{\infty} \frac{e^{-t}-1}{t^a}\,dt &= -\int_0^1 \int_0^{\infty} t^{1-a}e^{-st}\,dt\,ds \tag{$u = st$}\\
&= -\int_0^1 s^{a-2}\int_0^{\infty} u^{1-a}e^{-u}\,du\,ds \\
&= - \int_0^1 s^{a-2}\, \Gamma(2-a)\,ds\\
&= \frac{1}{1-a}\Gamma(2-a)\\
&= \Gamma(1-a).
\end{align}
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t =
\Gamma\pars{1 - a}\,,\qquad 1 < a < 2}$.



*

*$\ds{\large\textsf{First Approach}}$:
\begin{align}
\color{#f00}{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t} & =
\int_{t\ =\ 0}^{t\ \to\ \infty}\pars{\expo{-t} - 1}
\,{\dd\pars{t^{1 - a}} \over 1 - a} =
-\,{1 \over 1 - a}\int_{0}^{\infty}t^{1 - a}\pars{-\expo{-t}}\,\dd t
\\[5mm] & =
{\Gamma\pars{2 - a} \over 1 - a} = \color{#f00}{\Gamma\pars{1 - a}}
\end{align}


Note that
  $\ds{t^{1 - a}\pars{\expo{-t} - 1} \bbox[8px,#dfd]{vanishes\ out}}$ in the limits $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$ because $\ds{1 < a <2}$.


 2. $\ds{\large\textsf{Second Approach}}$:\begin{align}
\color{#f00}{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t} & =
\int_{0}^{\infty}\pars{\expo{-t} - 1}\ \overbrace{%
{1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1}\expo{-tx}
\,\dd x}^{\ds{1 \over t^{a}}}\ \,\dd t
\\[5mm] & =
{1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1}
\int_{0}^{\infty}\bracks{\expo{-\pars{x + 1}t} -\expo{-xt}}\,\dd t\,\dd x
\\[5mm] & =
{1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1}
\pars{{1 \over x + 1} - {1 \over x}}\,\dd x =
-\,{1 \over \Gamma\pars{a}}\int_{0}^{\infty}{x^{a - 2} \over x + 1}\,\dd x
\\[5mm] & \stackrel{1/\pars{x + 1}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over \Gamma\pars{a}}\int_{1}^{0}x\pars{{1 \over x} - 1}^{a - 2}
\,\pars{-\,{\dd x \over x^{2}}}
\\[5mm] & =
-\,{1 \over \Gamma\pars{a}}\int_{0}^{1}x^{1 - a}\pars{1 - x}^{a - 2}\,\dd x
\quad\pars{\begin{array}{l}\mbox{The integral is the}\ Beta\ Function
\\
\ds{\mrm{B}\pars{2 - a,a - 1}} \end{array}}
\\[2mm] & =
-\,{1 \over \Gamma\pars{a}}\,
{\Gamma\pars{2 - a}\Gamma\pars{a - 1} \over \Gamma\pars{1}}\quad
\pars{\begin{array}{l}\ds{\mrm{B}\pars{\mu,\nu} = {\Gamma\pars{\mu}\Gamma\pars{\nu} \over\Gamma\pars{\mu + \nu}}}
\\[2mm]
\ds{\Gamma}:\ Gamma\ Function
\end{array}}
\\[2mm] & =
-\,{1 \over \pi/\braces{\Gamma\pars{1 -a}\sin\pars{\pi\bracks{1 - a}}}}\,
\pi\sin\pars{\pi\bracks{a - 1}}\quad
\pars{\begin{array}{l}\mbox{We used twice}
\\
Euler\ Reflection
\\
Formula
\\[2mm]
\mbox{and}\ \ds{\Gamma\pars{1} = 1}\end{array}}
\\ & = \color{#f00}{\Gamma\pars{1 - a}}
\end{align}
