Question Number 74503 by crystal0207 last updated on 25/Nov/19
Commented by mathmax by abdo last updated on 25/Nov/19
$$\left.{a}\right)\int_{\mathrm{0}} ^{\infty} \:{x}^{\alpha−\mathrm{1}} \:{e}^{−\lambda{x}} {dx}\:=_{\lambda{x}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{{t}}{\lambda}\right)^{\alpha−\mathrm{1}} \:{e}^{−{t}} \:\frac{{dt}}{\lambda} \\ $$$$=\frac{\mathrm{1}}{\lambda^{\alpha} }\int_{\mathrm{0}} ^{\infty} \:\:{t}^{\alpha−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:=\frac{\mathrm{1}}{\lambda^{\alpha} }×\Gamma\left(\alpha\right)\:\:\left(\:\:\lambda>\mathrm{0}\right) \\ $$$$\left.{b}\right)\:\Gamma\left(\alpha+\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{x}^{\alpha} \:{e}^{−{x}} \:{dx}\:\:{and}\:{by}\:{parts}\:\:{u}={x}^{\alpha} \:{and}\:{v}^{'} \:={e}^{−{x}} \\ $$$$\Gamma\left(\alpha+\mathrm{1}\right)\:=\left[−{x}^{\alpha} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:\:\alpha{x}^{\alpha−\mathrm{1}} \:{e}^{−{x}} \:{dx} \\ $$$$=\alpha\Gamma\left(\alpha\right) \\ $$$$\left.{c}\right)\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\lambda{x}} \:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}+\mathrm{1}−\mathrm{1}} {e}^{−\lambda{x}} {dx}\:=\frac{\Gamma\left({n}+\mathrm{1}\right)}{\lambda^{{n}+\mathrm{1}} } \\ $$$$\Gamma\left({n}+\mathrm{1}\right)={n}\Gamma\left({n}−\mathrm{1}\right)\:={n}\left({n}−\mathrm{1}\right)\Gamma\left({n}−\mathrm{2}\right)={n}!\Gamma\left(\mathrm{1}\right) \\ $$$$\Gamma\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}\:} {dx}\:=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\lambda{x}} \:{dx}\:=\frac{{n}!}{\lambda^{{n}+\mathrm{1}} } \\ $$