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Evaluate $ \int_0^\infty\frac{dx}{x^2-2x+4}. $ I cannot figure it out. Any hint is appreciated.

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    Sorry, I corrected the problem. I think I can do it now.2012-12-05

2 Answers 2

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No need for complex integration:

$x^2-2x+4=(x-1)^2+3=3\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)\Longrightarrow$

$\int_0^\infty\frac{dx}{x^2-2x+4}=\frac{1}{\sqrt 3}\int_0^\infty\frac{\frac{1}{\sqrt 3}dx}{\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)}=$

$=\left.\frac{1}{\sqrt 3}\arctan\frac{x-1}{\sqrt 3}\right|_0^\infty=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\left(-\frac{\pi}{6}\right)\right)=\frac{2\pi}{3\sqrt 3}$

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    Thank you so much, @DonAntonio.2012-12-05
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Just in a nicer form: $\int_{0}^{+\infty}\frac{dx}{x^2-2x+4}=\int_{-1}^{+\infty}\frac{dx}{x^2+3}=\frac{1}{\sqrt{3}}\int_{-1/\sqrt{3}}^{+\infty}\frac{dz}{z^2+1}=\frac{1}{\sqrt{3}}\left(\frac{\pi}{2}+\arctan\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(\pi-\arctan\sqrt{3}\right)=\frac{2\pi}{3\sqrt{3}}.$

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    Thank you, Jack, your solution is also very good and innovative.2012-12-07