1-calculate-A-n-1-n-1-ln-1-x-2-1-x-2-dx-with-n-integr-and-n-1-2-find-lim-n-A-n-3-study-the-convergence-of-A-n-

Question Number 56329 by maxmathsup by imad last updated on 14/Mar/19

1) c a l c u l a t e A_{n} = \int_{\frac{1}{n}}^{1} \frac{l n (1 + x^{2})}{1 + x^{2}} d x w i t h n i n t e g r a n d n ⩾ 1

2) f i n d {l i m}_{n \to + \infty} A_{n}

3) s t u d y t h e c o n v e r g e n c e o f Σ A_{n}

Commented by maxmathsup by imad last updated on 17/Mar/19

1) A_{n} =_{x = t a n θ} \int_{a r c t a n (\frac{1}{n})}^{\frac{π}{4}} \frac{l n (1 + {t a n}^{2} θ)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ

= \int_{a r c t a n (\frac{1}{n})}^{\frac{π}{4}} l n (\frac{1}{{c o s}^{2} θ}) d θ = - 2 \int_{a r c t a n (\frac{1}{n})}^{\frac{π}{4}} l n (c o s θ) d θ \Rightarrow

{l i m}_{n \to + \infty} A_{n} = - 2 \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ l e t I = \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ a n d

J = \int_{0}^{\frac{π}{4}} l n (s i n θ) d θ w e h a v e I + J = \int_{0}^{\frac{π}{4}} l n (c o s θ s i n θ) d θ

= \int_{0}^{\frac{π}{4}} l n (\frac{s i n (2 θ)}{2}) d θ = - \frac{π}{4} l n (2) + \int_{0}^{\frac{π}{4}} l n (s i n (2 θ)) d θ b u t

\int_{0}^{\frac{π}{4}} l n (s i n (2 θ) d θ =_{2 θ = t} \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n t) d t = \frac{1}{2} (- \frac{π}{2} l n (2)) = - \frac{π}{4} l n (2) \Rightarrow

I + J = - \frac{π}{2} l n (2)

I = \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ =_{θ = t - \frac{π}{2}} \int_{\frac{π}{2}}^{\frac{3 π}{4}} l n (s i n θ) d θ = \int_{\frac{π}{2}}^{0} l n (s i n θ) d θ + \int_{0}^{\frac{3 π}{4}} l n (s i n θ) d θ

= \frac{π}{2} l n (2) + \int_{0}^{\frac{3 π}{4}} l n (s i n θ) d θ \dots . b e c o n t i n u e d \dots .