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Question Number 217855 by yamane last updated on 22/Mar/25

i n e e d h e l p

\int_{0}^{1} x^{n} (e^{- \frac{x^{2}}{2}}) d x, (n \in N)

Answered by mr W last updated on 23/Mar/25

i f n = 0 :

I_{0} = \int_{0}^{1} e^{- \frac{x^{2}}{2}} d x = \sqrt{\frac{π}{2}} e r f (\frac{1}{\sqrt{2}})

i f n = 1 :

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

= {[e^{- \frac{x^{2}}{2}}]}_{1}^{0} = 1 - \frac{1}{\sqrt{e}}

I_{n} = \int_{0}^{1} x^{n} e^{- \frac{x^{2}}{2}} d x

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

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

= \frac{1}{n + 1} (\frac{1}{\sqrt{e}} + I_{n + 2})

\Rightarrow I_{n + 2} = (n + 1) I_{n} - \frac{1}{\sqrt{e}}

I_{n} = (n - 1) I_{n - 2} - \frac{1}{\sqrt{e}}

= (n - 1) (n - 3) I_{n - 4} - (n - 3) \frac{1}{\sqrt{e}}

= \dots .

= (n - 1) (n - 3) \dots 3 \times 1 I_{0} - \frac{1}{\sqrt{e}} [1 + 3!! + 5!! + \dots + (n - 3)!!]

I_{n} = \sqrt{\frac{π}{2}} e r f (\frac{1}{\sqrt{2}}) (n - 1)!! - \frac{1}{\sqrt{e}} \underset{k = 1}{\sum^{n - 3}} k!! f o r n = e v e n

I_{n} = (1 - \frac{1}{\sqrt{e}}) (n - 1)!! - \frac{1}{\sqrt{e}} \underset{k = 2}{\sum^{n - 3}} k!! f o r n = o d d

Answered by Frix last updated on 22/Mar/25

\underset{0}{\int^{1}} x^{n} e^{- \frac{x^{2}}{2}} d x \overset{[t = \frac{x^{2}}{2}]}{=} 2^{\frac{n}{2} - \frac{1}{2}} \underset{0}{\int^{\frac{1}{2}}} t^{\frac{n}{2} - \frac{1}{2}} e^{- t} d t =

Let n = 2 z - 1

= 2^{z - 1} \underset{0}{\int^{\frac{1}{2}}} t^{z - 1} e^{- t} d t =

This is the l o w e r i n c o m p l e t e G a m m a f u n c t i o n

= 2^{z - 1} γ (z, \frac{1}{2}) = \frac{1}{{(2 e)}^{z}} \underset{k = 0}{\sum^{\infty}} \frac{1}{2^{k} \underset{j = 0}{\prod^{k}} (z + j)}; z = \frac{n + 1}{2}

\dots not nice \dots

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