91Ó°ÊÓ

The following information is known as exponential integrals.

$$\begin{gathered} E_n\left(x\right)=\underset{1}{\overset{∞}{∫}}\frac{e^{-xt}}{t^n}dt \\ Ei\left(x\right)=\underset{-∞}{\overset{t}{∫}}\frac{e^t}{t}dt \end{gathered}$$

The exponential integral is defined as follows.

$$\begin{gathered} {\mathit{E}}_{\mathbf{n}}\left(x\right)\mathbf{=}\underset{\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∫}}}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{x}\mathbf{t}}}{{\mathbf{t}}^{\mathbf{n}}}\mathit{d}\mathit{t} \\ \mathit{E}\mathit{i}\left(x\right)\mathbf{=}\underset{\mathbf{-}\mathbf{∞}}{\overset{\mathbf{t}}{\mathbf{∫}}}\frac{{\mathbf{e}}^{\mathbf{t}}}{\mathbf{t}}\mathit{d}\mathit{t} \end{gathered}$$

(a)

Make the following substitutions.

$\begin{array}{rcl}u& =& xt\\ du& =& xdt\end{array}$

Plug these in the integral and evaluate the expression.

$\begin{array}{rcl}{E}_n\left(x\right)& =& \underset{x}{\overset{∞}{∫}}\frac{e^{-u}{x}^n}{u^n}\frac{1}{x}du\\ & =& \underset{x}{\overset{∞}{∫}}{x}^{n-1}\frac{e^{-u}{x}^n}{u^n}du\\ & =& \underset{x}{\overset{∞}{∫}}\frac{e^{-u}}{u^n}du\end{array}$

(b)

Make the following substitutions.

$\begin{array}{rcl}t& =& -u\\ dt& =& -du\end{array}$

Plug these in the integral and evaluate the expression.

$\begin{array}{rcl}{E}_i\left(x\right)& =& \underset{∞}{\overset{-x}{∫}}\frac{e^{-u}}{-u}\left(-du\right)\\ & =& -\underset{-x}{\overset{∞}{∫}}\frac{e^{-u}}{-u}du\end{array}$

(c)

Evaluate the negative inside the domain.

$\begin{array}{rcl}{E}_i\left(-x\right)& =& -\underset{x}{\overset{∞}{∫}}\frac{e^{-u}}{u}du\\ & =& E_i\left(x\right)\end{array}$

(d)

Begin with the following expression.

$\begin{array}{rcl}{E}_i\left(-\frac{1}{x}\right)& =& -\underset{-\frac{1}{x}}{\overset{∞}{∫}}\frac{e^{-u}}{u}du\end{array}$

Make the following substitutions.

$\begin{array}{rcl}u& =& -\frac{1}{t}\\ du& =& \frac{1}{t^2}dt\end{array}$

Plug these in the integral and evaluate the expression.

$\begin{array}{rcl}\underset{x}{\overset{0}{∫}}\frac{e^{-{\displaystyle \frac{1}{u}}}u}{u^2}du& =& \underset{0}{\overset{x}{∫}}\frac{e^{{\displaystyle \frac{1}{u}}}}{u}du\end{array}$

Hence proved.