Q:

How can prove that $\int_{0}^{\infty}\sum_{i=0}^{j}(-1)^{i +j}{j\choose i}{e^{-x^{i +a}}\over x}dx={j!\gamma\over a(a+1)(a+2)\cdots(a+j)}?$

Where $\gamma=0.5772156...$ is the Euler's constant and $j$ is an integer, $j\ge 1.$
$a\in \Re$
How can I show that

$$\int_{0}^{\infty}\sum_{i=0}^{j}(-1)^{i +j}{j\choose i}{e^{-x^{i +a}}\over x}dx=-(-1)^j\color{red}{j!\gamma\over a(a+1)(a+2)\cdots(a+j)}?\tag1$$

Let take a simple case, $a=1$ and $j=1$
$$I=\int_{0}^{\infty}{e^{-x}-e^{-x^2}\over x}dx=-\color{blue}{\gamma\over 2}\tag2$$
Substitution $u=\ln{x}\rightarrow xdu=dx$
$x=\infty\rightarrow u=\infty$ and $x=0\rightarrow u=-\infty$
$$I=\int_{-\infty}^{\infty}e^{-e^u}-e^{-e^{2u}}du=-{\gamma\over 2}\tag3$$
I am not sure how to integrate $(3)$. Help needed, thank you. 

A:

We write $I(a,j)=\int^\infty_0\sum_{i=0}^{j}(-1)^{i}\binom{j}{i}\frac{\exp (-x^{a+i})}{x}\,dx$. 
Using $\binom{j}{i}=\binom{j-1}{i}+\binom{j-1}{i-1}$, we conclude $I(a,j)=I(a,j-1)-I(a+1,j-1)$.
If we can prove $I(a,1)=\frac{-\gamma}{a(a+1)}$, the expected form for $I(a,j)$ follows from induction. 
$
\begin{align*}
I(a,1)&=\int^\infty_0\frac{\exp(x^{-a})-\exp (-x^{a+1})}{x}\,dx\\
&=\lim_{z\to0}\int^\infty_z\frac{\exp(x^{-a})-\exp (-x^{a+1})}{x}\,dx\\
&=\lim_{z\to0}\left(\int^\infty_z\frac{\exp(x^{-a})}{x}\,dx-\int^\infty_z\frac{\exp (-x^{a+1})}{x}\,dx\right)\\
&=\lim_{z\to 0}\left(-\frac{Ei(-z^a)}{a}+\frac{Ei(-z^{a+1})}{a+1}\right)\\
&=\lim_{z\to 0}\left(-\frac{\gamma+\log(z^a)+O(z^a)}{a}+\frac{\gamma+\log(z^{a+1})+O(z^{a+1})}{a+1}\right)\\
&=\frac{-\gamma}{a(a+1)}. 
\end{align*}$ 

A:

Hint $$\color{red}{\int_{-\infty }^{\infty
 }{\frac{{{e}^{-{{x}^{\,\alpha }}}}-{{e}^{-{{x}^{\,\beta 
 }}}}}{x}}dx=\gamma \frac{\alpha -\beta }{\alpha \beta }}$$

Proof
Set $t=x^\alpha$ , $n=\frac{\beta}{\alpha}$, so $dx=\large {\frac{1}{\alpha x^{\alpha-1}}}$$dt$ and
$$I=\int_{-\infty }^{\infty
 }{\frac{{{e}^{-{{x}^{\,\alpha }}}}-{{e}^{-{{x}^{\,\beta 
 }}}}}{x}}dx=\frac{1}{\alpha}\int_{0}^{\infty}\frac{e^{-t}-e^{-t^{\large n}}}{t} dt$$
therefore
$$\qquad I=\frac{1}{\alpha }\left[ \left( \int_{0}^{1}{\frac{{{e}^{-t}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-t}}-1}{t}dt}} \right)-\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt}} \right) \right.  $$
$$=\frac{1}{\alpha }\left[ \left( \int_{0}^{1}{\frac{{{e}^{-t}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-t}}}{t}dt}} \right)-\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}} \right) \right].$$
Indeed, we have 
$$I=-\frac{1}{\alpha }\left[ \color{red}\gamma +\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}} \right) \right]\quad (1)$$
set 
$$J=\int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}}$$
and let $u=t^n$, then
$$J=\frac{1}{n}\int_{0}^{1}{\frac{{{e}^{-u}}-1}{u}du+\int_{1}^{\infty }{\frac{{{e}^{-u}}}{u}du}}=-\frac{1}{n}\color{red}\gamma\qquad\quad(2)$$
$(1)$ and $(2)$
$$I=-\frac{1}{\alpha }( \color{red}\gamma -\frac{1}{n}\color{red}\gamma)=-\frac{1}{\alpha }\color{red}\gamma\left(1-\frac{\alpha}{\beta}\right)=\color{red}{\gamma \frac{\alpha -\beta }{\alpha \beta }}$$

Note $$\color{blue}{\gamma
 =\int_{0}^{1}{\frac{1-{{e}^{-y}}}{y}dy-\int_{1}^{\infty }{\frac{{{e}^{-y}}}{y}dy}}}\qquad\qquad\qquad\qquad(3)$$

A:

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$\ds{j = 1,2,3,\ldots}$. With
  $\ds{\epsilon > 0\ \mbox{and}\ \ul{a > 0},\ }$ lets consider

\begin{align}
&\color{#f00}{\int_{\epsilon}^{\infty}
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} =
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}
\int_{\epsilon}^{\infty}{\expo{-x^{k +a}} \over x}\,\dd x\tag{1}
\end{align}
and
\begin{align}
\fbox{$\ds{\ \int_{\epsilon}^{\infty}{\expo{-x^{k +a}} \over x}\,\dd x\ }$} &\
\stackrel{x^{k + a}\phantom{aa}\mapsto\ x}{=}\
\int_{\epsilon}^{\infty}{\expo{-x} \over x^{1/\pars{k + a}}}
\,{1 \over k + a}\,x^{1/\pars{k + a} - 1}\,\dd x =
{1 \over k + a}\int_{\epsilon^{k + a}}^{\infty}{\expo{-x} \over x}\,\dd x
\\[3mm] & =
{1 \over k + a}\bracks{%
-\pars{k + a}\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} +
\int_{\epsilon^{k + a}}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}
\\[3mm] & =
-\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} +
{1 \over k + a}\int_{\epsilon^{k + a}}^{\infty}\ln\pars{x}\expo{-x}\,\dd x
\\[3mm] & =
-\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} +
{1 \over k + a}\ \underbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}
_{\ds{-\gamma}}\ -\
{1 \over k + a}\int_{0}^{\epsilon^{k + a}}\ln\pars{x}\expo{-x}\,\dd x
\end{align}

In the $\ds{\epsilon \to 0^{+}}$ limit, we'll have
$\ds{\pars{~\mbox{see expression }\ \pars{1}~}}$
\begin{align}
&\color{#f00}{\int_{0}^{\infty}
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} =
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}\pars{-\,{\gamma \over k + a}}
\tag{2}
\\[3mm] &\
\mbox{because}\quad
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k} = \delta_{j0}\quad
\mbox{and}\quad j \not= 0
\end{align}

$\pars{2}$ becomes
\begin{align}
&\color{#f00}{\int_{0}^{\infty}
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} =
-\gamma\,\pars{-1}^{\,j}\sum_{k = 0}^{j}\pars{-1}^{k}{j \choose k}
\int_{0}^{1}x^{k + a - 1}\,\dd x
\\[3mm] = &\
-\gamma\,\pars{-1}^{\,j}\int_{0}^{1}x^{a - 1}
\sum_{k = 0}^{j}{j \choose k}\pars{-x}^{k}\,\dd x =
-\gamma\,\pars{-1}^{\,j}\int_{0}^{1}x^{a - 1}\pars{1 - x}^{j}\,\dd x
\\[3mm] = &\
-\gamma\,\pars{-1}^{\,j}\,
{\Gamma\pars{a}\Gamma\pars{j + 1} \over \Gamma\pars{a + j + 1}} =
-\gamma\,\pars{-1}^{\,j}\,{j!  \over \pars{a}_{j + 1}}
\end{align}
where $\ds{\pars{a}_{j + 1}}$ is a Pochammer Symbol:
$$
\pars{a}_{j + 1} = a\pars{a +1}\pars{a + 2}\ldots\pars{a + j}
$$

Finally,
$$
\color{#f00}{\int_{0}^{\infty}
\sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x}
=
\color{#f00}{%
-\gamma\,\pars{-1}^{\,j}\,{j!  \over a\pars{a +1}\pars{a + 2}\ldots\pars{a + j}}}
$$

