Question Number 74503 by crystal0207 last updated on 25/Nov/19
Commented by mathmax by abdo last updated on 25/Nov/19
![a)∫_0 ^∞ x^(α−1) e^(−λx) dx =_(λx=t) ∫_0 ^∞ ((t/λ))^(α−1) e^(−t) (dt/λ) =(1/λ^α )∫_0 ^∞ t^(α−1) e^(−t) dt =(1/λ^α )×Γ(α) ( λ>0) b) Γ(α+1) =∫_0 ^∞ x^α e^(−x) dx and by parts u=x^α and v^′ =e^(−x) Γ(α+1) =[−x^α e^(−x) ]_0 ^∞ +∫_0 ^∞ αx^(α−1) e^(−x) dx =αΓ(α) c)∫_0 ^∞ x^n e^(−λx) dx =∫_0 ^∞ x^(n+1−1) e^(−λx) dx =((Γ(n+1))/λ^(n+1) ) Γ(n+1)=nΓ(n−1) =n(n−1)Γ(n−2)=n!Γ(1) Γ(1)=∫_0 ^∞ e^(−x ) dx =[−e^(−x) ]_0 ^(+∞) =1 ⇒∫_0 ^∞ x^n e^(−λx) dx =((n!)/λ^(n+1) )](https://www.tinkutara.com/question/Q74541.png)
$$\left.{a}\right)\int_{\mathrm{0}} ^{\infty} \:{x}^{\alpha−\mathrm{1}} \:{e}^{−\lambda{x}} {dx}\:=_{\lambda{x}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{{t}}{\lambda}\right)^{\alpha−\mathrm{1}} \:{e}^{−{t}} \:\frac{{dt}}{\lambda} \\ $$$$=\frac{\mathrm{1}}{\lambda^{\alpha} }\int_{\mathrm{0}} ^{\infty} \:\:{t}^{\alpha−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:=\frac{\mathrm{1}}{\lambda^{\alpha} }×\Gamma\left(\alpha\right)\:\:\left(\:\:\lambda>\mathrm{0}\right) \\ $$$$\left.{b}\right)\:\Gamma\left(\alpha+\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{x}^{\alpha} \:{e}^{−{x}} \:{dx}\:\:{and}\:{by}\:{parts}\:\:{u}={x}^{\alpha} \:{and}\:{v}^{'} \:={e}^{−{x}} \\ $$$$\Gamma\left(\alpha+\mathrm{1}\right)\:=\left[−{x}^{\alpha} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:\:\alpha{x}^{\alpha−\mathrm{1}} \:{e}^{−{x}} \:{dx} \\ $$$$=\alpha\Gamma\left(\alpha\right) \\ $$$$\left.{c}\right)\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\lambda{x}} \:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}+\mathrm{1}−\mathrm{1}} {e}^{−\lambda{x}} {dx}\:=\frac{\Gamma\left({n}+\mathrm{1}\right)}{\lambda^{{n}+\mathrm{1}} } \\ $$$$\Gamma\left({n}+\mathrm{1}\right)={n}\Gamma\left({n}−\mathrm{1}\right)\:={n}\left({n}−\mathrm{1}\right)\Gamma\left({n}−\mathrm{2}\right)={n}!\Gamma\left(\mathrm{1}\right) \\ $$$$\Gamma\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}\:} {dx}\:=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\lambda{x}} \:{dx}\:=\frac{{n}!}{\lambda^{{n}+\mathrm{1}} } \\ $$