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Question Number 34221 by abdo imad last updated on 03/May/18

$$studytheconvergenceof{\int }_0^1\frac{\sqrt{1-x}}{x}dx.$$

Commented by math khazana by abdo last updated on 04/May/18

$$letputx={sin}^2\theta \Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{cos\theta }{{sin}^2\theta }.2sin\theta cos\theta d\theta$$$$={\int }_0^{\frac{\pi }{2}}\frac{2{cos}^2\theta }{sin\theta }d\theta =2{\int }_0^{\frac{\pi }{2}}\frac{1-{sin}^2\theta }{sin\theta }d\theta$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{sin\theta }-2{\int }_0^{\frac{\pi }{2}}sin\theta d\theta butwehave$$$${\int }_0^{\frac{\pi }{2}}sin\theta d\theta ={[-cos\theta ]}_0^{\frac{\pi }{2}}=1andch.tan\left(\frac{\theta }{2}\right)=xgive$$$${\int }_{\xi }^{\frac{\pi }{2}}\frac{d\theta }{sin\theta }={\int }_{\xi }^{\frac{\pi }{4}}\frac{1}{\frac{2x}{1+x^2}}\frac{2dx}{1+x^2}={\int }_{\xi }^{\frac{\pi }{4}}\frac{dx}{x}$$$$=ln\left(\frac{\pi }{4}\right)-ln\xi so{lim}_{\xi \to 0^{+}}{\int }_{\xi }^{\frac{\pi }{4}}\frac{d\theta }{sin\theta }=+\mathrm{\infty }$$$$\Rightarrow Iisdivergent.$$

Answered by Joel578 last updated on 03/May/18

$$I=\underset{t\to 0^{+}}{\lim }\left({\int }_t^1\frac{\sqrt{1-x}}{x}dx\right)$$$$$$$${\int }_t^1\frac{\sqrt{1-x}}{x}dx(u=\sqrt{1-x}\to du=-\frac{1}{2u}dx)$$$$=-{\int }_{\sqrt{1-t}}^0\frac{2u^2}{1-u^2}du=2{\int }_{\sqrt{1-t}}^0\frac{u^2}{u^2-1}du$$$$=2{\int }_{\sqrt{1-t}}^0(1+\frac{1}{u^2-1})du$$$$=2{\int }_{\sqrt{1-t}}^0(1+\frac{1}{2(u-1)}-\frac{1}{2(u+1)})du$$$$=2{[u+\frac{1}{2}\ln \left(\frac{u-1}{u+1}\right)]}_{\sqrt{1-t}}^0$$$$=2[0+\frac{1}{2}\ln (-1)-\sqrt{1-t}-\frac{1}{2}\ln \left(\frac{\sqrt{1-t}-1}{\sqrt{1-t}+1}\right)]$$$$=i\pi -2\sqrt{1-t}-\ln \left(\frac{\sqrt{1-t}-1}{\sqrt{1-t}+1}\right)$$$$$$$$I=\underset{t\to 0^{+}}{\lim }[i\pi -2\sqrt{1-t}-\ln \left(\frac{\sqrt{1-t}-1}{\sqrt{1-t}+1}\right)]$$$$=i\pi -2-\ln 0$$$$=\mathrm{\infty }$$$$\mathrm{The}\mathrm{integral}\mathrm{doesnt}\mathrm{converge}$$

Commented by math khazana by abdo last updated on 04/May/18

$$sirjoelfromwherecomethecomplexi\pi the$$$$integralisreal....$$

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