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Question Number 31056 by abdo imad last updated on 02/Mar/18

$$find{\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2-2xcos\alpha +1}with0<\alpha <\pi .$$

Commented by abdo imad last updated on 03/Mar/18

$$I={\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2-2xcos\alpha +{cos}^2\alpha +{sin}^2\alpha }$$$$={\int }_1^{+\mathrm{\infty }}\frac{dx}{{(x-cos\alpha )}^2+{sin}^2\alpha }thech.x-cos\alpha =sin\alpha tgive$$$$I={\int }_{\frac{1-cos\alpha }{sin\alpha }}^{+\mathrm{\infty }}\frac{sin\alpha dt}{{sin}^2\alpha (1+t^2)}=\frac{1}{sin\alpha }{\int }_{tan\left(\frac{\alpha }{2}\right)}^{+\mathrm{\infty }}\frac{dt}{1+t^2}$$$$I=\frac{1}{sin\alpha }{\left[arctant\right]}_{tan\left(\frac{\alpha }{2}\right)}^{+\mathrm{\infty }}=\frac{1}{sin\alpha }(\frac{\pi }{2}-\frac{\alpha }{2})\Rightarrow$$$$I=\frac{\pi -\alpha }{2sin\alpha }.$$

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