let-I-1-dt-1-t-2-find-lim-0-I-

Question Number 36181 by prof Abdo imad last updated on 30/May/18

l e t I (ξ) = \int_{ξ}^{1 - ξ} \frac{d t}{1 - {(t - ξ)}^{2}}

f i n d {l i m}_{ξ \to 0^{+}} I (ξ)

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

c h a n g e m e n t t - ξ = s i n α g i v e α = a r c s i n (t - ξ) \Rightarrow

I (ξ) = \int_{0}^{a r c s i n (1 - 2 ξ)} \frac{c o s α d α}{1 - {s i n}^{2} α} = \int_{0}^{a r c s i n (1 - 2 ξ)} \frac{d α}{c o s α}

=_{t a n (\frac{α}{2}) = u} \int_{0}^{t a n (\frac{a r c s i n (1 - 2 ξ)}{2})} \frac{1}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{t a n (\frac{a r c s i n (1 - 2 ξ)}{2})} \frac{2 d u}{1 - u^{2}}

= \int_{0}^{t a n (\frac{a r c s i n (1 - 2 ξ)}{2})} (\frac{1}{1 + u} + \frac{1}{1 - u}) d u = {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{t a n (\frac{a r c s i n (1 - 2 ξ)}{2})}

= l n ∣ \frac{1 + t a n (\frac{a r c s i n (1 - 2 ξ)}{2})}{1 - t a n (\frac{a r c s i n (1 - 2 ξ)}{2})} ∣ \Rightarrow {l i m}_{ξ \to 0^{+}} I (ξ)

= l n ∣ \frac{1 + t a n (\frac{π}{4})}{1 - t a n (\frac{π}{4})} ∣ = + \infty .