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Proof of Equation (2.7)

\begin{equation*} {{I}^{(n)}}\left( c,k \right)=\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{\left( -1 \right)}^{n-1+i}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right]{{I}^{(0)}}\left( c,k+i \right) \right\}} \end{equation*}
Proof:
By definition, {{I}^{\left( 0 \right)}}\left( c,k \right)=\int{{{c}^{k}}\exp \left( c \right)dc}. Therefore,
\begin{equation*} {{I}^{\left( 0 \right)}}\left( c,k \right)={{c}^{k+1}}\exp \left( c \right)h_{k}^{c}. \end{equation*}

Using integration by parts, we can solve {{I}^{\left( 1 \right)}}\left( c,k \right)
\begin{align*} & {{I}^{\left( 1 \right)}}\left( c,k \right)\\ &=\int{\frac{{{I}^{\left( 0 \right)}}\left( c,k \right)}{c}dc} \\ & =\int{{{c}^{k}}\exp \left( c \right)h_{k}^{c}dc} \\ & =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\int{\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k+1}^{c}dc} \\ & =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+1}^{c}+\int{\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+2}^{c}dc} \\ & =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+1}^{c}+\frac{{{c}^{k+3}}}{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}\exp \left( c \right)h_{k+2}^{c}-\cdots \\ & =\frac{{{I}^{\left( 0 \right)}}\left( c,k \right)}{0!(k+1)}-\frac{{{I}^{\left( 0 \right)}}\left( c,k+1 \right)}{0!\left( k+1 \right)\left( k+2 \right)}+\frac{{{I}^{\left( 0 \right)}}\left( c,k+2 \right)}{0!\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}-\cdots. \end{align*}

By the same way, we derive
\begin{align*} {{I}^{\left( 2 \right)}}\left( c,k \right)=\frac{1}{1!{{\left( k+1 \right)}^{2}}}{{I}^{\left( 0 \right)}}\left( c,k \right)&-\Bigg[ \frac{1}{1!{{\left( k+1 \right)}^{2}}\left( k+2 \right)}\\&+\frac{1}{1!\left( k+1 \right){{\left( k+2 \right)}^{2}}} \Bigg]{{I}^{\left( 0 \right)}}\left( c,k+1 \right)+\cdots. \end{align*}

Repeating the same operations for n times, we derive
\begin{align*} {{I}^{(n)}}\left( c,k \right)&=\frac{{{\left( -1 \right)}^{n-1}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)} \right]{{I}^{(0)}}\left( c,k \right) \\&+\frac{{{\left( -1 \right)}^{n}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)\left( k+2 \right)} \right]{{I}^{(0)}}\left( c,k+1 \right) \\ & +\frac{{{\left( -1 \right)}^{n+1}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)} \right]{{I}^{(0)}}\left( c,k+2 \right) \\ & + \cdots \\ &=\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{\left( -1 \right)}^{n-1+i}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right]{{I}^{(0)}}\left( c,k+i \right) \right\}}. \end{align*}

\square

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