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Question Number 36728 by a1bgt3@gmail.com last updated on 04/Jun/18

the improper integral ∫_0 ^1 (dx/(√(1−x^2 ))) converges to

$${the}\:{improper}\:{integral}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{converges}\:{to} \\ $$

Commented by abdo.msup.com last updated on 05/Jun/18

I =lim_(ξ→0)  ∫_ξ ^1    (dx/(√(1−x^2 )))  but the changrment  x =sinθ  give    ∫_ξ ^1    (dx/(√(1−x^2 ))) = ∫_(arcsin(ξ)) ^(π/2)   ((cosθ dθ)/(cosθ))dθ  =(π/2) −arcsin(ξ) ⇒lim_(ξ→0) ∫_ξ ^1   (dx/(√(1−x^2 ))) =(π/2)  so I =(π/2) .

$${I}\:={lim}_{\xi\rightarrow\mathrm{0}} \:\int_{\xi} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{but}\:{the}\:{changrment} \\ $$$${x}\:={sin}\theta\:\:{give}\: \\ $$$$\:\int_{\xi} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:\int_{{arcsin}\left(\xi\right)} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\theta\:{d}\theta}{{cos}\theta}{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:−{arcsin}\left(\xi\right)\:\Rightarrow{lim}_{\xi\rightarrow\mathrm{0}} \int_{\xi} ^{\mathrm{1}} \:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\frac{\pi}{\mathrm{2}} \\ $$$${so}\:{I}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$

Answered by MJS last updated on 04/Jun/18

∫_0 ^1 (dx/(√(1−x^2 )))=[arcsin x]_0 ^1 =(π/2)  ...nothing improper about this...

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\left[\mathrm{arcsin}\:{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$$$...\mathrm{nothing}\:\mathrm{improper}\:\mathrm{about}\:\mathrm{this}... \\ $$

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