Question

We have encountered the gamma function$\Gamma \left(\alpha \right)$in our study of Bessel functions in Section 6.4 (page 263). One definition of this function is given by the improper integral

$\Gamma \left(\alpha \right)={\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt,\alpha >0.$

Use this definition to show that $\Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right).$ When $\alpha =n$ is a positive integer the last property can be used to show that$\Gamma (n+1)=n!$. See Appendix A.

Short answer

$\Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right)$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $t⩾0$ . Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Use gamma function

Using the definition of the gamma function

$\Gamma \left(\alpha \right)={\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt$

localid="1663915716650" $$\begin{gathered} \Gamma (\alpha +1)={\int }_0^{\infty }{t}^{(\alpha +1)-1}{e}^{-t}dt \\ ={\int }_0^{\infty }{t}^{\alpha }{e}^{t}dt \end{gathered}$$

Using the integration by parts.

$$\begin{gathered} v=e^{-t} \\ dv=-e^{t}dt \\ u=t^{\alpha }\backslash \mathrm{hfill} \\ du=\alpha t^{\alpha -1}dt, \end{gathered}$$

$$\begin{gathered} \Gamma (\alpha +1)={\int }_0^{\infty }{t}^{\alpha }{e}^{-t}dt \\ =-{\left.t^{\alpha }{e}^t\right|}_0^{\infty }-\underset{0}{\overset{\infty }{\int }}-\alpha t^{\alpha -1}{e}^{-t}dt \end{gathered}$$

Evaluate the integral. Reduce as necessary

$\Gamma (\alpha +1)=\underset{N\to \infty }{lim}-{\left.t^{\alpha }{e}^t\right|}_0^N-{\int }_0^{\infty }-e^{-t}\alpha t^{\alpha -1}dt$

localid="1663915847732" $$\begin{gathered} =\underset{N\to \infty }{lim}-N^{\alpha }{e}^{-N}-\left(-0^{\alpha }{e}^{-0}\right)+\alpha {\int }_0^{\infty }{t}^{\alpha -1}{e}^{t}dt \\ =0+\alpha {\int }_{\mathit{0}}^{\infty }{t}^{\alpha \mathit{-}\mathit{1}}{e}^{t}dt \end{gathered}$$

Here note that${\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt$is the definition of$\Gamma \left(\alpha \right)$

$\Gamma (\alpha +1)=\alpha {\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt$

$\therefore \Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right)$