Question

(a) Express E₁(x)as an incomplete$\mathit{\Gamma }$function.

(b) Find the asymptotic series for E₁(x).

Short answer

1. The required expression is, $E_1\left(x\right)=\underset{x}{\overset{\infty }{\int }}{t}^{-1}{e}^{-t}dt=\Gamma \left(0,x\right)$.
2. The required series is, $E_1\left(x\right)=\frac{e^{-x}}{x}\left[1-\frac{1}{x}+\frac{2}{x^2}-\frac{6}{x^3}+...\right]$.

Step-by-step solution

Given information

E₁(x) and gamma function is given.

Definition of a Gamma function

The Gamma Function is defined as

$\mathit{\Gamma }\left(p\right)\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{p}\mathbf{>}\mathbf{0}$

Begin with the definition of a Gamma function

(a)

Write the Gamma function.

$\Gamma \left(\mathrm{p}\right)={\int }_0^{\infty }{x}^{\mathrm{p}-1}{e}^{-\mathrm{x}}dx,p>0$

Substitute p = 0.

$\begin{array}{rcl}\Gamma (0,\mathrm{x})& =& {\int }_0^{\infty }{t}^{-1}{e}^{-\mathrm{x}}d\\ & =& E_1\left(x\right)\end{array}$

Write the expansion of an incomplete Gamma function.

(b)

Expand the incomplete Gamma function.

$\Gamma (p,x)=x^{p-1}{e}^{-x}\left[1+\frac{\left(p-1\right)}{x}+\frac{\left(p-1\right)\left(p-2\right)}{x^2}+\frac{\left(p-1\right)\left(p-2\right)\left(p-3\right)}{x^3}+...\right]$

Substitute p = 0.

$\begin{array}{rcl}\Gamma (0,x)& =& E_1\left(x\right)\\ & =& \frac{e^{-x}}{x}\left[1-\frac{1}{x}+\frac{2}{x^2}-\frac{6}{x^3}+...\right]\end{array}$

Hence, the asymptotic series is $\begin{array}{rcl}& & \frac{e^{-x}}{x}\left[1-\frac{1}{x}+\frac{2}{x^2}-\frac{6}{x^3}+...\right]\end{array}$.