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Question Number 5835 by sanusihammed last updated on 31/May/16

E v a l u a t e t h e i n t e g r a l .

\int [(x - \frac{x^{3}}{2} + \frac{x^{5}}{2.4} - \frac{x^{7}}{2.4.6} + \dots) (1 - \frac{x^{2}}{2^{2}} + \frac{x^{4}}{2^{2} . 4^{2}} - \frac{x^{6}}{2^{2} . 4^{2} . 6^{2}} + \dots .)] d x

f o r 0 < x < \infty

P l e a s e h e l p

Commented by Yozzii last updated on 31/May/16

\frac{1}{(2 n - 2)!!} = \frac{1}{2 \times 4 \times 6 \times 8 \times 10 \times \dots \times (2 n - 2)}

= \frac{1}{2^{n - 1} (1 \times 2 \times 3 \times 4 \times 5 \times \dots \times (n - 1))}

∴ \frac{1}{(2 n)!!} = \frac{1}{2^{n} n!} (n ⩾ 0)

(2 n)!! = 2 n {(2 n - 2)!!}

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 1}}{2^{n} (n!)} = \sqrt{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 1}}{{(\sqrt{2})}^{2 n + 1} n!}

= x \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} {(\frac{x^{2}}{2})}^{n}

= x \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} {(- \frac{x^{2}}{2})}^{n}

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 1}}{2^{n} (n!)} = x e x p (\frac{- x^{2}}{2}) = - \frac{d}{d x} (e^{- x^{2} / 2})

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1 - \frac{x^{2}}{2^{2}} + \frac{x^{4}}{2^{2} 4^{2}} - \frac{x^{6}}{2^{2} 4^{2} 6^{2}} + \dots = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{{(2^{n} n!)}^{2}}

= \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n!)}^{2}} {(- \frac{x^{2}}{4})}^{n} = \underset{n = 0}{\sum^{\infty}} {\frac{1}{n!} {(\frac{- x}{2})}^{n}} {\frac{1}{n!} {(\frac{x}{2})}^{n}}

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∴ J = \int {x e x p (- 0.5 x^{2}) (\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{{(2^{n} n!)}^{2}})} d x

J = \underset{n = 0}{\sum^{\infty}} [\frac{{(- 1)}^{n}}{{(2^{n} n!)}^{2}} {\int ({x e}^{- 0.5 x^{2}}) x^{2 n} d x}]

= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2^{n} n!)}^{2}} {- e^{- 0.5 x^{2}} x^{2 n} + 2 n \int {x e}^{- 0.5 x^{2}} x^{2 n - 2} d x}

= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2^{n} n!)}^{2}} (- x^{2 n} e^{- 0.5 x^{2}} - 2 {n x}^{2 n - 2} e^{- 0.5 x^{2}} - 2 n (2 n - 2) x^{2 n - 4} e^{- 0.5 x^{2}} + 2 n (2 n - 2) (2 n - 4) \int x^{2 n - 6} {x e}^{- 0.5 x^{2}} d x)

= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{((2 n)!!)}^{2}} (- e^{- 0.5 x^{2}} (x^{2 n} + 2 {n x}^{2 (n - 1)} + 2 n (2 n - 2) x^{2 (n - 2)} + 2 n (2 n - 2) (2 n - 4) x^{2 (n - 3)} + \dots . + (2 n)!! x^{2 (n - n)})) + C

J = \underset{n = 0}{\sum^{\infty}} (\frac{{(- 1)}^{n + 1} e^{- 0.5 x^{2}}}{{((2 n)!!)}^{2}} {\underset{k = 0}{\sum^{n}} x^{2 (n - k)} \frac{(2 n)!!}{(2 (n - k))!!}}) + C

J = e^{- 0.5 x^{2}} \underset{n = 0}{\sum^{\infty}} (\frac{{(- 1)}^{n + 1}}{(2 n)!!} {\underset{k = 0}{\sum^{n}} \frac{1}{(2 (n - k))!!} x^{2 (n - k)}}) + C

J = - e^{- 0.5 x^{2}} \underset{n = 0}{\sum^{\infty}} (\frac{1}{n!} {(\frac{- x^{2}}{2})}^{n} {\underset{k = 0}{\sum^{n}} \frac{1}{(n - k)!} {(\frac{x^{2}}{2})}^{n - k}})

J = - e^{- 0.5 x^{2}} \underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} {\frac{1}{n! (n - k)!} {(\frac{- x^{2}}{2})}^{n} {(\frac{x^{2}}{2})}^{n - k}}

J = - e^{- 0.5 x^{2}} \underset{n = 0}{\sum^{\infty}} (\frac{1}{n!} {(\frac{- x^{2}}{2})}^{n}) (\underset{k = 0}{\sum^{\infty}} \frac{1}{(n - k)!} {(\frac{x^{2}}{2})}^{n - k}) + C ? ? ?

J = - e^{- 0.5 x^{2}} e^{- 0.5 x^{2}} e^{0.5 x^{2}} + C

J = - e^{- 0.5 x^{2}} + C

J ∣_{0}^{\infty} = - e^{- 0.5 x^{2}} ∣_{0}^{\infty} = 1 .

Commented by sanusihammed last updated on 31/May/16

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