Question Number 36181 by prof Abdo imad last updated on 30/May/18
$${let}\:{I}\left(\xi\right)\:\:=\:\int_{\xi} ^{\mathrm{1}−\xi} \:\:\:\frac{{dt}}{\mathrm{1}−\left({t}−\xi\right)^{\mathrm{2}} } \\ $$$${find}\:{lim}_{\xi\rightarrow\mathrm{0}^{+} } \:\:\:{I}\left(\xi\right) \\ $$
Commented by maxmathsup by imad last updated on 20/Aug/18
![changement t−ξ =sinα give α =arcsin(t−ξ) ⇒ I(ξ) = ∫_0 ^(arcsin(1−2ξ)) ((cosα dα)/(1−sin^2 α)) =∫_0 ^(arcsin(1−2ξ)) (dα/(cosα)) =_(tan((α/2))=u) ∫_0 ^(tan(((arcsin(1−2ξ))/2))) (1/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 )) = ∫_0 ^(tan(((arcsin(1−2ξ))/2))) ((2du)/(1−u^2 )) = ∫_0 ^(tan(((arcsin(1−2ξ))/2))) ((1/(1+u)) +(1/(1−u)))du =[ln∣((1+u)/(1−u))∣]_0 ^(tan(((arcsin(1−2ξ))/2))) =ln∣ ((1+tan(((arcsin(1−2ξ))/2)))/(1−tan(((arcsin(1−2ξ))/2))))∣ ⇒lim_(ξ→0^+ ) I(ξ) =ln∣ ((1 +tan((π/4)))/(1−tan((π/4))))∣ =+∞ .](https://www.tinkutara.com/question/Q42238.png)
$${changement}\:\:{t}−\xi\:={sin}\alpha\:{give}\:\:\alpha\:={arcsin}\left({t}−\xi\right)\:\Rightarrow \\ $$$${I}\left(\xi\right)\:=\:\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)} \:\:\:\:\frac{{cos}\alpha\:{d}\alpha}{\mathrm{1}−{sin}^{\mathrm{2}} \alpha}\:=\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)} \:\:\frac{{d}\alpha}{{cos}\alpha} \\ $$$$=_{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)={u}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)} \:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)} \:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)} \left(\frac{\mathrm{1}}{\mathrm{1}+{u}}\:+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du}\:=\left[{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\right]_{\mathrm{0}} ^{{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)} \\ $$$$={ln}\mid\:\frac{\mathrm{1}+{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{arcsin}\left(\mathrm{1}−\mathrm{2}\xi\right)}{\mathrm{2}}\right)}\mid\:\Rightarrow{lim}_{\xi\rightarrow\mathrm{0}^{+} } \:\:\:{I}\left(\xi\right) \\ $$$$={ln}\mid\:\frac{\mathrm{1}\:+{tan}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}−{tan}\left(\frac{\pi}{\mathrm{4}}\right)}\mid\:\:=+\infty\:. \\ $$