Evaluate-the-following-integral-0-n-x-1-x-dx-n-N-Here-x-is-the-integer-part-of-x-e-g-0-12-0-5-896-5-

Question Number 1444 by 112358 last updated on 04/Aug/15

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

\int_{0}^{n} ⌊ x ⌋^{1 / ⌊ x ⌋!} d x (n \in N)

H e r e ⌊ x ⌋ i s t h e i n t e g e r - p a r t o f x

e . g ⌊ 0.12 ⌋ = 0, ⌊ 5.896 ⌋ = 5

Answered by 123456 last updated on 05/Aug/15

f (x) = ⌊ x ⌋^{1 / ⌊ x ⌋!} \equiv n^{1 / n!}, x = n + r, x \in R_{+}, n \in N, r \in [0, 1)

I_{n} = \underset{0}{\int^{n}} ⌊ x ⌋^{1 / ⌊ x ⌋!} d x

I_{0} = 0

I_{n} = \underset{0}{\int^{n}} ⌊ x ⌋^{1 / ⌊ x ⌋!} d x = \underset{i = 0}{\sum^{n - 1}} \underset{i}{\int^{i + 1}} i^{1 / i!} d x

= \underset{i = 0}{\sum^{n - 1}} i^{1 / i!}

Commented by 123456 last updated on 06/Aug/15

L = \lim_{n \to + \infty} n^{1 / n!} = ?

D = \lim_{n \to + \infty} n^{1 / n}

\ln D = \lim_{n \to + \infty} \frac{\ln n}{n} \overset{L . H}{=} \lim_{n \to 0} \frac{1}{n} = 0

D = 1

\ln L = \lim_{n \to + \infty} \frac{\ln n}{n!} = ?

0 < n^{1 / n!} < n^{1 / n} for largd enought n

0 < \frac{\ln n}{n!} \overset{?}{<} \frac{\ln n}{n}

1 / n! > 0

1 < n^{1 / n!} < n^{1 / n} for large enought n

L = 1

Commented by 123456 last updated on 05/Aug/15

\lim_{n \to + \infty} n^{1 / n!} \neq 0

\underset{i = 0}{\sum^{+ \infty}} i^{1 / i!} diverge

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