Question

The logarithmic integralis $\mathit{l}\mathit{i}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{\int }}}\frac{\mathbf{dt}}{\mathbf{In}\mathbf{}\mathbf{t}}$. Express as exponential integrals

1. $\mathit{l}\mathit{i}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$
   
2. $\mathit{l}\mathit{i}\mathbf{\left(}{\mathbf{e}}^{\mathbf{x}}\mathbf{\right)}$
   
3. $\mathit{l}\mathit{i}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{\int }}}\frac{\mathbf{dt}}{\mathbf{In}\mathbf{\left(}{\displaystyle \frac{1}{t}}\mathbf{\right)}}$
   

Short answer

The following required expressions are shown:

1. $li\left(x\right)=E_i\left(Inx\right)$
   
2. $li\left(e^x\right)=E_i\left(x\right)$
   
3. $=-E_i\left(Inx\right)$
   

Step-by-step solution

Given information

The definition of a logarithmic integral is given.

$li\left(\mathrm{x}\right)=\underset{0}{\overset{\mathrm{x}}{\int }}\frac{\mathrm{dt}}{\mathrm{In}\mathrm{t}}$

Begin with making a substitution.

(a)

Make the following substitution.

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

Apply the substitution in the given integrand.

$\begin{array}{rcl}li\left(x\right)& =& \underset{-\infty }{\overset{In\left(x\right)}{\int }}\frac{e^u}{u}du\\ & =& E_i\left(Inx\right)\end{array}$

Substitute the values in the domain.

(b)

Substitute e^x in the domain of li(x).

$\begin{array}{rcl}li\left(e^x\right)& =& \underset{-\infty }{\overset{x}{\int }}\frac{e^u}{u}du\\ & =& E_i\left(x\right)\end{array}$

Repeat the process.

(c)

Use the following substitution.

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

Apply the substitution in the given integrand.

$\begin{array}{rcl}\underset{-\infty }{\overset{In\mathit{}x}{\int }}\frac{e^u}{-u}du& =& \underset{In\left(x\right)}{\overset{-\infty }{\int }}\frac{e^u}{u}du\\ & =& -E_i\left(Inx\right)\end{array}$

Thus, the following expressions are shown:

$$\begin{gathered} \left(a\right)li\left(x\right)=E_i(Inx) \\ \left(b\right)li\left(e^x\right)=E_i\left(x\right) \\ \left(c\right)=-E_i(Inx) \end{gathered}$$