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Question Number 203301 by MathematicalUser2357 last updated on 15/Jan/24

is

\int \sqrt{x} e^{- x} d x

= \underset{n = 1}{\sum^{\infty}} ((\underset{k = 1}{\prod^{n}} \frac{2}{2 k + 1}) x^{n} \sqrt{x} e^{- x}) + C ?

Answered by witcher3 last updated on 15/Jan/24

\int x^{a} e^{- x} dx = \frac{x^{n + 1}}{n + 1} e^{- x} + \frac{1}{n + 1} \int x^{n + 1} e^{- x}

I_{n} = \frac{x^{n + 1} e^{- x}}{n + 1} + \frac{1}{n + 1} I_{n + 1}

I_{n} = \frac{1}{n + 1} (x^{n + 1} e^{- x} + \frac{1}{n + 2} (x^{n + 2} e^{- x} + \frac{1}{n + 3} (x^{n + 3} e^{- x} + \frac{1}{n + 4} I_{n + 4} + \dots .)

\sum_{m ⩾ 1} (\underset{k = 1}{\prod^{m}} \frac{1}{(n + k)}) x^{n + m} e^{- x} + c

\int \sqrt{x} e^{- x} dx; n = \frac{1}{2}

= \sum_{n ⩾ 1} (\underset{k = 1}{\prod^{n}} \frac{1}{(\frac{1}{2} + k)}) x^{\frac{1}{2} + n} e^{- x} = \sum_{n ⩾ 1} (\underset{k = 1}{\prod^{n}} \frac{2}{k + 2}) x^{n} \sqrt{x} e^{- x} + c

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