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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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The integral is divergent so it doesn't exist, no matter how you try to calculate it. –  DonAntonio Dec 5 '12 at 23:01
    
Sorry, I corrected the problem. I think I can do it now. –  Sam Dec 5 '12 at 23:06
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2 Answers

up vote 4 down vote accepted

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. –  Sam Dec 5 '12 at 23:20
    
Any time......... –  DonAntonio Dec 5 '12 at 23:22
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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. –  Sam Dec 7 '12 at 3:24
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