lim-n-0-1-log-1-sin-n-x-1-x-x-n-dx-

Question Number 147344 by mathdanisur last updated on 20/Jul/21

\underset{n \to \infty}{l i m} \underset{0}{\int^{1}} l o g (\frac{1 + {s i n}^{n} x}{1 + x + x^{n}}) d x = ?

Answered by mathmax by abdo last updated on 20/Jul/21

f_{n} (x) = \log (\frac{1 + \sin^{n} x}{1 + x + x^{n}}) \to cs to f (x) = \log (\frac{1}{1 + x}) = - \log (1 + x) \Rightarrow

\lim_{n \to + \infty} \int_{0}^{1} f_{n} (x) dx = \int_{0}^{1} \lim_{n \to + \infty} f_{n} (x) dx = - \int_{0}^{1} \log (1 + x) dx

=_{1 + x = t} - \int_{1}^{2} \log (t) dt = - {[tlot - t]}_{1}^{2} = - (2 \log 2 - 2 + 1)

= 1 - 2 \log 2

Commented by mathdanisur last updated on 20/Jul/21

t h a n k y o u S e r

Commented by mathmax by abdo last updated on 20/Jul/21

you are welcome sir

Commented by mathdanisur last updated on 20/Jul/21

T h a n k s S e r, b u t h o w d i d y o u c h a n g e

(i n t e r c h a n g e o f) t h e l i m i t t o t h e

i n t e g r a l, p l e a s e n o t e S e r \dots

Commented by mathmax by abdo last updated on 21/Jul/21

theoreme de convergence dominee \dots

Commented by mathdanisur last updated on 21/Jul/21

S e r, c a n y o u s t a t e t h a t t h e o r e m, p l e a s e