Question

The integral $\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{t}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathit{d}\mathit{t}\mathbf{=}\mathit{\Gamma }\mathbf{\text{'}}\left(p,x\right)$is called an incomplete $\mathit{\Gamma }\mathbf{\text{'}}$function. [Note that if x = 0, this integral is$\mathit{\Gamma }\mathbf{\text{'}}\mathbf{\left(}\mathit{p}\mathbf{\right)}$.] By repeated integration by parts, find several terms of the asymptotic series for$\mathit{\Gamma }\mathbf{\text{'}}\mathbf{(}\mathit{p}\mathbf{,}\mathit{x}\mathbf{)}$.

Short answer

The several terms of asymptotic series is mentioned below.$\Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}\left(1+\frac{\left(p-1\right)}{x}+\frac{\left(p-1\right)\left(p-2\right)}{x^2}+\frac{\left(p-1\right)\left(p-2\right)\left(p-3\right)}{x^3}+...\right)$.

Step-by-step solution

Given Information

The incomplete gamma function is$\underset{0}{\overset{\infty }{\int }}{t}^{p-1}{e}^{-t}dt=\Gamma \text{'}(p,x)$.

Definition of Gamma function.

Gamma function behaves similarly to a factorial for natural numbers (a discrete set), its extension to positive real numbers (a continuous set) allows it to be used to model situations involving continuous change.

Gamma function$\mathit{\Gamma }\mathbf{\text{'}}\mathbf{\left(}\mathit{p}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}$.

Find the terms of asymptotic series.

The incomplete gamma function is $\underset{0}{\overset{\infty }{\int }}{t}^{p-1}{e}^{-t}dt=\Gamma \text{'}(p,x)$.

Partially integrate the function mentioned above.

The function becomes as follows.

$$\begin{gathered} \Gamma \text{'}(p.x)={\left(-t^{p-1}{e}^{-t}\right)}_x^{\infty }+\left(p-1\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt \\ \Gamma \text{'}(p.x)=x^{p-1}{e}^{-x}+\left(p-1\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt \end{gathered}$$

Integrate partially $\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt$.

The integration becomes as follows.

$\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt=t^{p-2}{e}^{-t}+\left(p-2\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-3}{e}^{-t}dt$

Substitute the value of integration mentioned above in the equation given below.

$$\begin{gathered} \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt \\ \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)\left({\left(-t^{p-2}{e}^{-t}\right)}_x^{\infty }+\left(p-2\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-3}{e}^{-t}dt\right) \\ \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)x^{p-2}{e}^{-x}+\left(p-1\right)\left(p-2\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-3}{e}^{-t}dt \end{gathered}$$

Integrate partially.

The integration becomes as follows.

$\underset{x}{\overset{\infty }{\int }}{t}^{p-2}{e}^{-t}dt=x^{p-3}{e}^{-x}+\left(p-3\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-4}{e}^{-t}dt$

Substitute the value of integration mentioned above in the equation given below.

$$\begin{gathered} \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)x^{\left(p-2\right)}{e}^{-x}+\left(p-1\right)\left(p-2\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-3}{e}^{-t}dt \\ \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)x^{p-2}{e}^{-x}+\left(p-1\right)\left(p-2\right)\left(x^{p-3}{e}^{-x}+\left(p-3\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-4}{e}^{-t}dt\right) \\ \Gamma \text{'}(p,x)=x^{p-1}{e}^{-x}+\left(p-1\right)x^{p-2}{e}^{-x}+\left(p-1\right)\left(p-2\right)x^{p-3}{e}^{-x}+\left(p-1\right)\left(p-2\right)\left(p-3\right)\underset{x}{\overset{\infty }{\int }}{t}^{p-4}{e}^{-t}dt \end{gathered}$$

Hence, the several terms of asymptotic series is mentioned below:

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