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Question Number 34222 by abdo imad last updated on 03/May/18

l e t g i v e t h e s e q u e n c e o f i n t e g r a l s

J_{n} = \int_{0}^{\infty} x^{n} e^{- \frac{x^{2}}{2}} d x

1) p r o v e t h a t J_{n} = (n - 1) J_{n - 2} \forall n ⩾ 2

2) c a l c u l a t e J_{2 p} a n d J_{2 p + 1} b y u s i n g f a c t o r i e l s .

3) p r o v e t h a t \forall n ⩾ 1 J_{n}^{2} ⩽ J_{n - 1} . J_{n + 1} .

4) p r o v e t h a t \frac{2^{2 p} {(p!)}^{2}}{(2 p)!} \frac{1}{\sqrt{2 p + 1}} ⩽ J_{0} ⩽ \frac{2^{2 p} {(p!)}^{2}}{(2 p)!} \frac{1}{\sqrt{2 p}}

5) f i n d a e q u i v a l e n t o f \frac{2^{2 p} {(p!)}^{2}}{(2 p)!} (p \to + \infty)

Commented by candre last updated on 03/May/18

J_{n} = \underset{0}{\int^{\infty}} x^{n} e^{- \frac{x^{2}}{2}} d x

u = x^{n - 1} \Rightarrow d u = (n - 1) x^{n - 2} d x

d v = {x e}^{- \frac{x^{2}}{2}} d x \Rightarrow v = - e^{- \frac{x^{2}}{2}}

= - {[x^{n - 1} e^{- \frac{x^{2}}{2}}]}_{0}^{\infty} + (n - 1) \int x^{n - 2} e^{- \frac{x^{2}}{2}} d x

= (n - 1) J_{n - 2}

\int {x e}^{- \frac{x^{2}}{2}} d x = - \int e^{u} d u = - e^{u} = - e^{- \frac{x^{2}}{2}}

u = - \frac{x^{2}}{2} \Rightarrow d u = - x d x

Commented by candre last updated on 03/May/18

J_{0} = \underset{0}{\int^{\infty}} e^{- \frac{x^{2}}{2}} d x

J_{0}^{2} = {(\underset{0}{\int^{\infty}} e^{- \frac{x^{2}}{2}} d x)}^{2} = \underset{0}{\int^{\infty}} e^{- x^{2} / 2} d x \underset{0}{\int^{\infty}} e^{- y^{2} / 2} d y

= \underset{0}{\int^{\infty}} \underset{0}{\int^{\infty}} e^{- \frac{x^{2} + y^{2}}{2}} d x d y

= \underset{0}{\int^{π / 2}} \underset{0}{\int^{\infty}} e^{- \frac{r^{2}}{2}} r d r d θ

= \underset{0}{\int^{π / 2}} d θ \underset{0}{\int^{\infty}} e^{- r^{2} / 2} r d r

u = - r^{2} / 2 \Rightarrow d u = - r d r

= \frac{π}{2} \cdot \underset{- \infty}{\int^{0}} e^{u} d u

= \frac{π}{2}

J_{0} = \sqrt{\frac{π}{2}}

Commented by candre last updated on 03/May/18

J_{2 p} :

J_{2} = 1 \cdot J_{0}

J_{4} = 3 \cdot J_{2} = 3 \cdot 1 \cdot J_{0}

J_{6} = 5 \cdot J_{4} = 5 \cdot 3 \cdot 1 \cdot J_{0}

J_{2 p} = (2 p - 1)!! J_{0}

J_{2 p + 1} :

J_{3} = 2 \cdot J_{1}

J_{5} = 4 \cdot J_{3} = 4 \cdot 2 \cdot J_{1}

J_{2 p + 1} = (2 p)!! J_{1} = 2^{p} p! J_{1}

Commented by abdo mathsup 649 cc last updated on 05/May/18

1) l e t p r o v e t h a t

J_{n + 1} = {n J}_{n - 1}

J_{n + 1} = \int_{0}^{\infty} x^{n + 1} e^{- \frac{x^{2}}{2}} d x

= - \int_{0}^{\infty} x^{n} (- {x e}^{- \frac{x^{2}}{2}} d x) = - ({[x^{n} e^{- \frac{x^{2}}{2}}]}_{0}^{+ \infty} - \int_{0}^{\infty} n x^{n - 1} e^{- \frac{x^{2}}{2}} d x)

= n \int_{0}^{\infty} x^{n - 1} e^{- \frac{x^{2}}{2}} d x = n J_{n - 1} \Rightarrow

J_{n} = (n - 1) J_{n - 2} \forall n ⩾ 2

2) J_{2 p} = (2 p - 1) J_{2 p - 2} \Rightarrow

\prod_{p = 1}^{n} J_{2 p} = \prod_{p = 1}^{n} J_{2 p - 2} \prod_{p = 1}^{n} (2 p - 1) \Rightarrow

J_{2} . J_{4} \dots . . J_{2 n} = 1.3.5 \dots . (2 n - 1) J_{0} . J_{2} \dots . J_{2 n - 2} \Rightarrow

J_{2 n} = 1.3.5 \dots . (2 n - 1) J_{0}

= \frac{1.2.3.4 \dots \dots (2 n - 1) (2 n)}{2^{n} n!} J_{0}

= \frac{(2 n)!}{2^{n} (n!)} J_{0}

J_{0} = \int_{0}^{\infty} e^{- \frac{x^{2}}{2}} d x =_{\sqrt{2} t = x} \int_{0}^{\infty} e^{- t^{2}} \sqrt{2} d t = \sqrt{2} \frac{\sqrt{π}}{2} = \sqrt{\frac{π}{2}}

\Rightarrow J_{2 n} = \frac{(2 n)!}{2^{n} n!} \sqrt{\frac{π}{2}}

J_{2 p + 1} = 2 p J_{2 p - 1} \Rightarrow \prod_{p = 1}^{n} J_{2 p + 1} = \prod_{p = 1}^{n} (2 p) \prod_{p = 1}^{n} J_{2 p - 1}

\Rightarrow J_{3} . J_{5} \dots . . J_{2 n + 1} = 2^{n} n! J_{1} . J_{3} \dots . J_{2 n - 1}

\Rightarrow J_{2 n + 1} = 2^{n} n! J_{1}

J_{1} = \int_{0}^{\infty} x e^{- \frac{x^{2}}{2}} d x = - \int_{0}^{\infty} (- x) e^{- \frac{x^{2}}{2}} d x

= - {[e^{- \frac{x^{2}}{2}}]}_{0}^{+ \infty} = 1 \Rightarrow J_{2 n + 1} = 2^{n} (n!) .