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Question Number 169436 by mathocean1 last updated on 30/Apr/22

Commented by mathocean1 last updated on 30/Apr/22

$$with0<\alpha <1$$

Answered by Mathspace last updated on 01/May/22

$$x=\frac{1}{t}\Rightarrow I={\int }_{\frac{1}{\alpha }}^{\alpha }\frac{-lnt}{1+\frac{1}{t^2}}(-\frac{dt}{t^2})$$$$=-{\int }_{\alpha }^{\frac{1}{\alpha }}\frac{lnt}{t^2+1}dt=-I\Rightarrow 2I=0\Rightarrow I=0$$

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