## 3.3 Rules Related to Derivative and Integration

The first partial derivative of $h_{s}^{c}$ with resepct to $c$ is
\begin{equation*}
\frac{\partial \left( h_{s}^{c} \right)}{\partial c}=\frac{-1}{\left( s+1 \right)\left( s+2 \right)}+\frac{-2{{\left( -c \right)}^{1}}}{\left( s+1 \right)\left( s+2 \right)\left( s+3 \right)}+\cdots ,
\end{equation*}
and it is equivalent to the first forward difference on $s$ \begin{equation*}
\frac{\partial \left( h_{s}^{c} \right)}{\partial c}=h_{s+1}^{c}-h_{s}^{c}.\tag{3.7}
\end{equation*}
[Proof for (3.7)]

Taking the $n$th partial derivative with respect to $c$, the result is the $n$th forward difference on $s$
\begin{equation*}
\frac{{{\partial }^{n}}\left( h_{s}^{c} \right)}{\partial {{c}^{n}}}=\Delta _{1}^{n}\left[ {{h}^{c}} \right]\left( s \right).\tag{3.8}
\end{equation*}
[Proof for (3.8)]

Applying (3.7) repeatedly, we can solve the first-order antiderivative with respect to $c$ as shown in (3.10). Without loss of generality, I have not included the constant in the presentation of antiderivatives throughout this paper. \begin{equation*}
\int{h_{s}^{c}dc=-\sum\limits_{i=0}^{\infty }{h_{s+i}^{c}}}. \tag{3.10}
\end{equation*}

[Proof for (3.10)]

The $n$th-order antiderivative with respect to $c$ is
\begin{equation*}
\int_{{}}^{(n)}h_{s}^{c}dc={{\left( -1 \right)}^{n}}\sum\limits_{i=0}^{\infty }{{i+n-1 \choose i}
h_{s+i}^{c}}. \tag{3.11}
\end{equation*}
[Proof for (3.11)]

Similarly,
\begin{equation*}
{\exp(c)}h_{s-n}^{c}=\int_{{}}^{\left( n \right)}{{\exp(c)}h_{s}^{c}dc}.
\end{equation*}
The result in (3.10) provides a key to solving the exponential integral. First, we take the Taylor expansion of $\exp \left( u \right)$ and integrate each term individually,
\begin{equation*}
\int{\frac{{\exp(u)}}{u}du}=\log\left|u\right|+\frac{{{u}^{1}}}{1!(1)}+\frac{{{u}^{2}}}{2!(2)}+\frac{{{u}^{3}}}{3!(3)}+\cdots .
\end{equation*}
Next, let
\begin{equation*}
y=\frac{{{u}^{1}}}{1!(1)}+\frac{{{u}^{2}}}{2!(2)}+\frac{{{u}^{3}}}{3!(3)}+\cdots,
\end{equation*}
and the first derivative of $y$ with respect to $u$ is $h_{0}^{-u}$. Then, $y$ can be derived from integrating $h_{0}^{-u}$ by $u$. According to (3.10), $y=\sum\limits_{s=0}^{\infty }{h_{s}^{-u}}$, and we solve the antiderivative of the exponential function as

\int{\frac{{\exp(u)}}{u}du}=\log\left| u \right|+\sum\limits_{s=0}^{\infty }{h_{s}^{-u}}. \tag{3.12}

[Examples of (3.12)]

As (2.2) indicates, the analytical solution of the exponential integral can be expressed as ${\exp(u)}h_{-1}^{u}$. From (3.12), we can derive the definition of $h_{-1}^{c}$ as (3.1) shows.

A useful identity that links $h_{s}^{c}$ and $h_{s+1}^{c}$ can be expressed as
\begin{equation*}
\left( s+1 \right)h_{s}^{c}=1-ch_{s+1}^{c}. \tag{3.13}
\end{equation*}
[Proof for (3.13)]

We can use this identity to deduce $h_{s}^{c}$ when $s$ is a negative integer smaller than $-1$. For example, let $s=-2$, and we derive $h_{-2}^{c}=-1+ch_{-1}^{c}$. Repeat the same identity by assuming $s=-3$, $s=-4$, and so on. The result completes the definition of $h_{s}^{c}$ for any real number $s$ and $c$ in (3.1).

In addition, the identity formula (2.2) makes both of the derivative and integral operators reversible:
\begin{align*}
&\frac{\partial \left( {{x}^{s}}{\exp(-x)} \right)}{\partial x}=s{{x}^{s-1}}{\exp(-x)}-{{x}^{s}}{\exp(-x)},\\
&\int{\left( s{{x}^{s-1}}{\exp(-x)}-{{x}^{s}}{\exp(-x)} \right)}dx={{x}^{s}}{\exp(-x)}\tag{3.14},
\end{align*}
where $sh_{s-1}^{-x}-xh_{s}^{-x}=1$.
[Proof for (3.14)]

The $n$th-order partial derivative with respect to $s$ can be generalized as
\begin{equation*}
\frac{{{\partial }^{\left( n \right)}}\left( h_{s}^{c} \right)}{\partial {{s}^{n}}}=\exp (-c)\sum\limits_{i=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}n!{{c}^{i}}}{i!{{\left( s+1+i \right)}^{n+1}}}}. \tag{3.15}
\end{equation*}
[Proof for (3.15)]

where ${{w}_{1}}=1$, ${{w}_{2}}=c{{w}_{1}}+\frac{d}{dc}\left( c{{w}_{1}} \right)$, $\cdots$, ${{w}_{n}}=c{{w}_{n-1}}+\frac{d}{dc}\left( c{{w}_{n-1}} \right)$.