Question Number 66404 by ~ À ® @ 237 ~ last updated on 14/Aug/19
Commented by mathmax by abdo last updated on 14/Aug/19
![let I_n =∫_0 ^∞ x^(2n) e^(−(x^2 /a^2 )) dx changement (x/a)=t (a>0) give I_n =∫_0 ^∞ (at)^(2n) e^(−t^2 ) (adt) =a^(2n+1) ∫_0 ^∞ t^(2n) e^(−t^2 ) dt and ∫_0 ^∞ t^(2n) e^(−t^2 ) dt =_(t=(√u)) ∫_0 ^∞ u^n e^(−u) (du/(2(√u))) =(1/2) ∫_0 ^∞ u^(n−(1/2)) e^(−u) du =(1/2)∫_0 ^∞ u^(n−1+(1/2)) e^(−u) du =(1/2)Γ(n+(1/2))⇒I_n =(a^(2n+1) /2)Γ(n+(1/2)) let W_n =∫_0 ^∞ t^(2n) e^(−t^2 ) dt by parts u^′ =t^(2n) and v=e^(−t^2 ) W_n =[(1/(2n+1))t^(2n+1) e^(−t^2 ) ]_0 ^∞ −∫_0 ^∞ (t^(2n+1) /(2n+1))(−2t)e^(−t^2 ) dt =(2/(2n+1))∫_0 ^∞ t^(2(n+1)) e^(−t^2 ) dt =(2/(2n+1)) W_(n+1) ⇒W_(n+1) =((2n+1)/2)W_n ⇒ Π_(k=0) ^(n−1) W_(k+1) =(1/2^n )Π_(k=0) ^(n−1) (2k+1)Π_(k=0) ^(n−1) W_n ⇒ W_1 .W_2 .....W_n =(1/2^n )(1.3.5.....(2n−1)W_0 .W_1 ....W_(n−1) ⇒ W_n =(1/2^n )(((2.3.4.5.6......(2n))/(2.4......(2n))))W_0 =(((2n)!)/(2^(2n) n!))((√π)/2) =((√π)/2^(2n+1) ) (((2n)!)/(n!)) ⇒ I_n =a^(2n+1) ((√π)/2^(2n+1) ) (((2n)!)/(n!)) =((a/2))^(2n+1) (((2n)!)/(n!)) (√π) also we get Γ(n+(1/2))=2W_n =((√π)/2^(2n) ) (((2n)!)/(n!))](https://www.tinkutara.com/question/Q66409.png)
$${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{2}{n}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} {dx}\:\:\:\:{changement}\:\frac{{x}}{{a}}={t}\:\:\:\:\:\:\:\:\:\:\left({a}>\mathrm{0}\right)\:{give} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\left({at}\right)^{\mathrm{2}{n}} \:{e}^{−{t}^{\mathrm{2}} } \left({adt}\right)\:={a}^{\mathrm{2}{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=_{{t}=\sqrt{{u}}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{u}^{{n}} \:{e}^{−{u}} \:\frac{{du}}{\mathrm{2}\sqrt{{u}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:{u}^{{n}−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{u}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:{u}^{{n}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{u}} \:{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Rightarrow{I}_{{n}} =\frac{{a}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}}\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${let}\:{W}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} {e}^{−{t}^{\mathrm{2}} } {dt}\:\:\:\:{by}\:{parts}\:\:{u}^{'} ={t}^{\mathrm{2}{n}} \:{and}\:{v}={e}^{−{t}^{\mathrm{2}} } \\ $$$${W}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\infty} \:−\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\left(−\mathrm{2}{t}\right){e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}\left({n}+\mathrm{1}\right)} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:{W}_{{n}+\mathrm{1}} \:\Rightarrow{W}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{W}_{{n}} \:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{W}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{W}_{{n}} \:\Rightarrow \\ $$$${W}_{\mathrm{1}} .{W}_{\mathrm{2}} …..{W}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{1}\right){W}_{\mathrm{0}} .{W}_{\mathrm{1}} ….{W}_{{n}−\mathrm{1}} \:\Rightarrow\right. \\ $$$${W}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\frac{\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}.\mathrm{6}……\left(\mathrm{2}{n}\right)}{\mathrm{2}.\mathrm{4}……\left(\mathrm{2}{n}\right)}\right){W}_{\mathrm{0}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!}\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\:\Rightarrow \\ $$$${I}_{{n}} ={a}^{\mathrm{2}{n}+\mathrm{1}} \:\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\:=\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\:\sqrt{\pi} \\ $$$${also}\:{we}\:{get}\:\:\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}{W}_{{n}} =\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2}{n}} }\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 14/Aug/19
$${Thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 14/Aug/19
$${you}\:{are}\:{welcome}. \\ $$