find-the-value-of-0-e-x-x-dx-

Question Number 27613 by abdo imad last updated on 10/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\infty} e^{- [x] - x} d x .

Commented by abdo imad last updated on 12/Jan/18

I_{n} = \sum_{k = 0}^{n} \int_{k}^{k + 1} e^{- [x] - x} d x I = l i m I_{n}

I_{n} = \sum_{k =}^{n} e^{- k} {[e^{- x}]}_{k}^{k + 1} = \sum_{k = 0}^{n} e^{- k} (e^{- k} - e^{- k - 1})

I_{n} = \sum_{k = 0}^{n} e^{- 2 k} - e^{- 1} \sum_{k = 0}^{n} e^{- 2 k} = (1 - e^{- 1}) \sum_{k = 0}^{n} {(e^{- 2})}^{k}

\int_{0}^{\infty} e^{- [x] - x} d x = {l i m}_{n - >\propto} I_{n} = (1 - e^{- 1}) . \frac{1}{1 - e^{- 2}}

= \frac{e^{2} (1 - e^{- 1})}{e^{2} - 1} = \frac{e^{2} - e}{e^{2} - 1} = \frac{e (e - 1)}{(e + 1) (e - 1)} = \frac{e}{e + 1} .

Answered by prakash jain last updated on 12/Jan/18

= \underset{i = 0}{\sum^{\infty}} e^{- i} \int_{i}^{i + 1} e^{- x} d x

= \underset{i = 0}{\sum^{\infty}} e^{- i} {[- e^{- x}]}_{i}^{i + 1}

= \underset{i = 0}{\sum^{\infty}} \frac{e^{- i}}{e^{i}} - \frac{e^{- i}}{e^{i + 1}}

= \underset{i = 0}{\sum^{\infty}} \frac{1}{e^{2 i}} - \frac{1}{e^{2 i + 1}}

= \underset{i = 0}{\sum^{\infty}} \frac{1}{e^{2 i}} - \underset{i = 0}{\sum^{\infty}} \frac{1}{e^{2 i + 1}}

= \frac{1}{1 - \frac{1}{e^{2}}} - \frac{1 / e}{1 - \frac{1}{e^{2}}}

= \frac{e (e - 1)}{e^{2} - 1} = \frac{e}{e + 1}