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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!))](Q66409.png)
Commented by ~ À ® @ 237 ~ last updated on 14/Aug/19
Commented by mathmax by abdo last updated on 14/Aug/19
