Question-45885

Question Number 45885 by Meritguide1234 last updated on 17/Oct/18

Commented by maxmathsup by imad last updated on 18/Oct/18

l e t A_{n} = \int_{0}^{\frac{π}{2}} n (1 - {(s i n x)}^{\frac{1}{n}}) d x = \int_{0}^{\frac{π}{2}} f_{n} (x) d x w i t h f_{n} (x) = n (1 - ({s i n x}^{\frac{1}{n}}))

b u t w e h a v e {(s i n x)}^{\frac{1}{n}} = e^{\frac{1}{n} l n (s i n x)} = 1 + \frac{l n (s i n x)}{n} + o (\frac{1}{n}) \Rightarrow

1 - e^{\frac{1}{n} l n (s i n x)} = - \frac{l n (s i n x)}{n} + o (\frac{1}{n}) \Rightarrow n (1 - e^{\frac{l n (s i n x)}{n}}) = - l n (s i n x) + o (1) \Rightarrow

f_{n} (x) = - l n (s i n x) + o (1) \Rightarrow {l i m}_{n \to + \infty} A_{n} = \int_{0}^{\frac{π}{2}} - l n (s i n x) d x

= - (- \frac{π}{2} l n (2)) = \frac{π}{2} l n (2) .

Commented by Meritguide1234 last updated on 19/Oct/18

n i c e a p p r o a c h

Commented by maxmathsup by imad last updated on 20/Oct/18

t h a n k y o u s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18

\lim_{n \to \infty} n \int_{0}^{\frac{π}{2}} {1 - {(s i n x)}^{\frac{1}{n}}} d x

\lim_{n \to \infty} n [∣ x ∣_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} {x c o z}^{2 q - 1} x d x]

h e r e 2 p - 1 = \frac{1}{n} 2 p = 1 + \frac{1}{n} p = \frac{n + 1}{2 n}

2 q - 1 = 0 q = \frac{1}{2} u s i n g g a m m a b e t a f u n c t i o n

f o r m u l a \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} {x c o s}^{2 a - 1} x d x = \frac{⌈ (p) ⌈ (q)}{2 ⌈ (p + q)}

\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (p) ⌈ (q)}{2 ⌈ (p + q)}]

\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (\frac{n + 1}{2 n}) ⌈ (\frac{1}{2})}{2 ⌈ (\frac{n + 1}{2 n} + \frac{1}{2})}]

\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (1 + \frac{1 - n}{2 n}) \times \sqrt{π}}{2 ⌈ (1 + \frac{n + 1}{2 n} - \frac{1}{2})}]

\lim_{n \to \infty} n [\frac{π}{2} - \frac{\sqrt{π} \times ⌈ (1 + \frac{1 - n}{2 n})}{2 ⌈ (1 + \frac{1}{2 n})}]

Commented by Meritguide1234 last updated on 18/Oct/18

n o t c o r r e c t

Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18

y e s i k n o w t h e r e s u l t i s y e t t o c o m e \dots w h e n n \to \infty

t h e r e s u l t s i s \infty

Commented by Meritguide1234 last updated on 18/Oct/18

p u t n = 1 / t \dots t h e n a n s i s (\frac{π}{2} l n 2)