Involving Euler's constant and Gamma function $$\int_{0}^{\infty}{e^{-x}+x-1\over x(e^{x/a}-1)}\mathrm dx=a\gamma+\ln{\Gamma(1+a)}\tag1$$
$\gamma$ is Euler-Mascheroni constant,

How can we show that $(1)=a\gamma+\ln{\Gamma(1+a)}?$

 A: Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{e^{-x}+x-1}{e^{x/a}-1}\frac{\mathrm{d}x}x
&=\frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x\\
&=\int_0^\infty\left(1-e^{-ax}\right)\frac{e^{-x}}{1-e^{-x}}\,\mathrm{d}x\\
&=\int_0^\infty\sum_{k=1}^\infty\left(e^{-kx}-e^{-(k+a)x}\right)\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a}\right)
\end{align}
$$
Since
$$
\left.\int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x\right|_{a=0}=0
$$
and by Gautschi's Inequality
$$
\frac{\Gamma(n+a+1)}{\Gamma(n+1)}=n^a\left(1+O\!\left(\frac1n\right)\right)
$$
we get
$$
\begin{align}
\int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac ak-\log\left(\frac{k+a}k\right)\right)\\
&=\lim_{n\to\infty}\left(aH_n-\log\left(\frac{\Gamma(n+a+1)}{\Gamma(a+1)\Gamma(n+1)}\right)\right)\\
&=\scriptsize\lim_{n\to\infty}\left(a\left(\log(n)+\gamma+O\!\left(\frac1n\right)\right)+\log(\Gamma(a+1))-a\log(n)+O\!\left(\frac1n\right)\right)\\[6pt]
&=a\gamma+\log(\Gamma(a+1))
\end{align}
$$
A: To evaluate this integral we will make use the following integral representation for the digamma function $\psi (x)$
$$\psi (x) = \int^\infty_0 \left (\frac{e^{-t}}{t} - \frac{e^{-xt}}{1 - e^{-t}} \right ) \, dt, \qquad x > 0. \tag1$$
Note that if we set $x = 1$ in the above integral representation for the digamma function, as $\psi (1) = -\gamma$ where $\gamma$ is the Euler–Mascheroni constant, we obtain the following integral representation for this constant 
$$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx, \tag2$$
and is the second result we intend to make use of.
Now, let
$$I = \int^\infty_0 \frac{e^{-x} + x - 1}{x(e^{x/a} - 1)} \, dx.$$
Setting $x \mapsto a x$ the integral becomes
$$I = \int^\infty_0 \frac{e^{-ax} + ax - 1}{x(e^x - 1)} \, dx.$$
Rearranging the numerator this can be rewritten
\begin{align*}
I &= \int^\infty_0 \frac{a e^{-x} + e^{-ax} - a e^{-x} + au + (a - 1) - a}{x(e^x - 1)} \, dx\\
&= a \int^\infty_0 \frac{e^{-x} + x - 1}{x(e^x - 1)} \, dx + \int^\infty_0 \frac{a e^{-ax} - a e^{-x} + a - 1}{x (e^x - 1)} \, dx. \tag3
\end{align*}
From (2), if we note that
$$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{xe^x} \right ) \, dx = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{e^{-x}}{x} \right ) \, dx = \int^\infty_0 \frac{x - 1 + e^{-x}}{x(e^x - 1)} \, dx,$$
the first integral appearing in (3) can be written in terms of the Euler-Mascheroni constant giving
\begin{align*}
I &= a \gamma + \int^\infty_0 \frac{e^{-ax} - a e^{-x} + a - 1}{x (e^x - 1)} \, dx,
\end{align*}
or
\begin{align*}
I &= a \gamma + \int^\infty_0 \frac{a (1 - e^{-x}) - (1 - e^{-ax})}{x (e^x - 1)} \, dx = a \gamma + \int^\infty_0 \left \{\frac{a}{x e^x} - \frac{1 - e^{-ax}}{x (e^x - 1)} \right \} \, dx, \tag4
\end{align*}
after rearranging the integrand.
To find the last integral that has appears, as
$$\psi (x) = \frac{d}{dx} \ln \Gamma (x),$$
then
$$\ln \Gamma (x) = \int^x_1 \psi (u) \, du.$$
From the integral representation for the digamma function, namely (1), we can write the above expression for $\ln \Gamma (x)$ as a double integral, namely
$$\ln \Gamma (x) = \int^\infty_0 \int^x_1 \left (\frac{e^{-t}}{t} - \frac{e^{-ut}}{1 - e^{-t}} \right ) \, du dt,$$
after the order of integration has been changed. The $u$-integration can be readily performed. Thus
\begin{align*}
\ln \Gamma (x) &= \int^\infty_0 \left [\frac{e^{-t}}{t} u + \frac{e^{-ut}}{t (1 - e^{-t})} \right ]^x_1 \, dt\\
&= \int^\infty_0 \left [(x - 1) - \frac{1 - e^{-(x - 1)t}}{1 - e^{-t}} \right ] \frac{e^{-t}}{t} \, dt.
\end{align*}
Now if $x \mapsto x + 1$, then
$$\ln \Gamma (x + 1) = \int^\infty_0 \left [\frac{x}{t e^t} - \frac{1 - e^{-xt}}{t(e^t - 1)} \right ] \, dt.$$
Thus the integral appearing in (4) is equal to $\Gamma (a + 1)$ and yields
$$\int^\infty_0 \frac{e^{-x} + x - 1}{x(e^{x/a} - 1)} \, dx = a \gamma + \ln \Gamma (a + 1),$$
as required.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}{\expo{-x} + x - 1 \over x\pars{\expo{x/a} - 1}}\,\dd x}} =
\int_{0}^{\infty}{\sum_{m = 2}^{\infty}\pars{-x}^{m}/m! \over
x\pars{1 - \expo{-x/a}}}\expo{-x/a}\,\dd x
\\ = &\
\sum_{m = 2}^{\infty}{\pars{-1}^{m} \over m!}
\int_{0}^{\infty}x^{m - 1}\sum_{n = 0}^{\infty}\expo{-\pars{n + 1}x/a}\,\dd x =
\sum_{m = 2}^{\infty}{\pars{-1}^{m} \over m!}\sum_{n = 0}^{\infty}\
\overbrace{\int_{0}^{\infty}x^{m - 1}\expo{-\pars{n + 1}x/a}\dd x}
^{\ds{\pars{m - 1}!\pars{a \over n + 1}^{m}}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}\
\underbrace{\sum_{m = 2}^{\infty}{1 \over m}\pars{-\,{a \over n + 1}}^{m}}
_{\ds{{a \over n + 1} - \ln\pars{1 + {a \over n + 1}}}}\ =
\lim_{N \to \infty}\sum_{n = 1}^{N}
\bracks{{a \over n} - \ln\pars{1 + {a \over n}}}
\\[5mm] = &\
a\ \underbrace{\lim_{N \to \infty}\pars{\sum_{n = 1}^{N}{1 \over n} - \ln\pars{N}}}_{\ds{=\ \gamma}}\ +\
\lim_{N \to \infty}\bracks{a\ln\pars{N} +
\sum_{n = 1}^{N}\ln\pars{n \over n + a}}
\\[5mm] = &\
a\gamma + \lim_{N \to \infty}\sum_{n = 1}^{N}\ln\pars{N^{a/N}n \over n + a} =
a\gamma + \lim_{N \to \infty}\ln\pars{\prod_{n = 1}^{N}{N^{a/N}n \over n + a}} =
a\gamma + \lim_{N \to \infty}\ln\pars{N^{a}\, N! \over
\bracks{1 + a}^{\,\overline{N}}}
\\[5mm] = &\
a\gamma + \lim_{N \to \infty}\ln\pars{N^{a}\, N! \over
\Gamma\pars{1 + a + N}/\Gamma\pars{1 + a}} =
a\gamma + \lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a}
\,{N^{a}\, N! \over \bracks{N + a}!}}
\\[5mm] = &\
a\gamma + \lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a}
\,{N^{a}\, \root{2\pi}N^{N + 1/2}\expo{-N} \over
\root{2\pi}\pars{N + a}^{N + a + 1/2}\expo{-\pars{N + a}}}}
\\[5mm] = &\
a\gamma +
\lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a}
\,{1 \over \pars{1 + a/N}^{N}\expo{-a}}} =
\bbx{a\gamma + \ln\pars{\Gamma\pars{1 + a}}}
\end{align}

because $\ds{\lim_{N \to \infty}\pars{1 + a/N}^{N} = \expo{a}}$ and
  $\ds{\lim_{N \to \infty}\pars{1 + a/N}^{a + 1/2} = 1}$.

